Shortcuts & Speed Methods

Number Types and Decimals

Tier 1 gives you 15 minutes for 25 maths questions, which is about 36 seconds a question. Number system questions are usually the quickest ones in the paper, so the aim is to finish them in 15 to 20 seconds and carry the spare time to the long questions. Most of the speed here is in checking instead of working: one test number like −1 breaks a false statement about number sets, and the last digit of the options often settles a product or a square root by itself. Under each method is the case where it goes wrong, for example using the terminating test before the fraction is in lowest terms.

Test one member that breaks the rule

When to use it: the options are statements or names of sets, like "Every whole number is a natural number" or "Which set contains only positive numbers?"

Fast way: don't try to prove a statement true. Try to break it with one number. Keep a small test kit in your head: 0, −1, 1/2 and $\sqrt{2}$. These four break almost every false statement in this topic.

For example: "Every integer is a whole number." Try −1. It is an integer but not a whole number, so the statement is false. That took three seconds.

Where it fails: a test number can only prove a statement false. If nothing in your kit breaks it, it is probably true, but read the definition once to be sure. Watch the small words too. "Rational × irrational is irrational" is broken by $0 \times \sqrt{2} = 0$, which is why the correct rule says non-zero rational.

Recurring decimal to fraction: the 9s and 0s rule

When to use it: a bar appears in the question, or you need to add or subtract recurring decimals.

Fast way: on top, write all the digits after the point minus the digits that don't repeat. Underneath, write one 9 for each repeating digit, then one 0 for each non-repeating digit.

For example: $0.3\overline{18} = \frac{318 - 3}{990} = \frac{315}{990} = \frac{7}{22}$

Where it fails: the whole-number part stays outside the rule, so $2.\overline{4} = 2 + \frac{4}{9} = \frac{22}{9}$, not $\frac{24}{99}$. The 9s always come before the 0s (990, never 099 or 909). And if nothing repeats there are no 9s at all: 0.35 is just 35/100.

Terminating test: only 2s and 5s

When to use it: "Which of these fractions has a terminating decimal?"

Fast way: cut the fraction to lowest terms, then look only at the denominator. If its prime factors are only 2s and 5s, the decimal terminates. The number of decimal places is the bigger of the two powers.

For example: 21/56 looks like it should recur, because 56 = 8 × 7. But 21/56 = 3/8 and 8 = $2^3$, so it terminates after 3 places: 0.375.

Where it fails: when you skip the reducing step. A 3 or a 7 in the denominator only matters if it is still there after cancelling. Once the fraction is in lowest terms, the numerator doesn't decide anything.

Divisibility: Composite Divisors and Missing Digits

Composite divisors: split into co-prime parts

When to use it: the divisor is 12, 18, 24, 36, 45, 72, 88, 99 or some other composite number.

Fast way: write the divisor as a product of two co-prime numbers whose rules you know, and check both.

For example: is 4,73,616 divisible by 24? The last three digits, 616, are 8 × 77, so the 8 test passes. The digit sum is 4 + 7 + 3 + 6 + 1 + 6 = 27, so the 3 test passes. Yes, it is divisible by 24.

For 99 there is an even quicker way. Split the number into pairs of digits from the right and add the pairs. If the total is a multiple of 99, so is the number. For 6,64,884: 84 + 48 + 66 = 198 = 2 × 99, so it is divisible by 99.

Where it fails: the two parts must be co-prime. 24 = 4 × 6 looks fine, but 4 and 6 share a 2. The number 36 is divisible by both 4 and 6, yet 36 ÷ 24 is not a whole number. The same thing goes wrong with 72 = 4 × 18: 36 passes both, but it is not a multiple of 72. The pair-adding trick works only for 99 and its factors, not for nearby numbers like 98 or 101.

Missing digits: the 9 rule and the 11 rule

When to use it: a digit is replaced by a letter and the number has to be divisible by 9, 11 or a composite like 72, 88 or 99.

Fast way (9): add the known digits. The missing digit is whatever lifts that sum to the next multiple of 9.

Fast way (11): add the digits in the odd places (counting from the right), add the digits in the even places, and make the difference 0, 11 or −11.

For example (9): 63k812 is divisible by 9. The known digits add to 6 + 3 + 8 + 1 + 2 = 20. The next multiple of 9 is 27, so k = 7.

For example (11): 8k5764 is divisible by 11. Odd places from the right: 4 + 7 + k = 11 + k. Even places: 6 + 5 + 8 = 19. The difference is k − 8, which must be 0, so k = 8.

With two missing digits and a divisor like 72 or 88, settle the 8 test first. It uses only the last three digits, so it pins down the last digit quickly. Then use the 9 rule or the 11 rule for the other one.

Where it fails:

• If the known digits already add up to a multiple of 9, the missing digit can be 0 or 9. "How many values are possible?" questions are built on exactly this.
• For 11, people try only a difference of 0. If that gives a digit below 0 or above 9, try 11 and −11 before giving up.
• In the 8 test, two last digits can both work (480 and 488 are both multiples of 8), so check each case.

Least Number to Add or Subtract

Use the remainder

When to use it: "What least number must be added to (or subtracted from) N to make it divisible by d?" The same idea gives the largest or smallest n-digit multiple of d.

Fast way: divide once and find the remainder r. To subtract, take away r. To add, put in d − r.

For example: 8,439 ÷ 37 leaves remainder 3, because 37 × 228 = 8,436. The least number to subtract is 3. The least number to add is 37 − 3 = 34.

For n-digit multiples, start from the edge. The largest 4-digit number, 9999, leaves 9 when divided by 37, so the largest 4-digit multiple of 37 is 9999 − 9 = 9990. The smallest 4-digit number, 1000, leaves 1, so the smallest 4-digit multiple is 1000 + (37 − 1) = 1036.

Where it fails:

• Mixing up the two answers. Subtract r, add d − r. The options nearly always contain both, so read the question again before you mark.
• If the remainder is 0, the answer is 0. Nothing needs to change.
• With several divisors ("divisible by 12, 18 and 30"), find their LCM first and use that as d.

Counting and Adding Multiples

How many multiples in a range

When to use it: "How many numbers from a to b are divisible by k?"

Fast way:

$$\text{count} = \left\lfloor \frac{b}{k} \right\rfloor - \left\lfloor \frac{a-1}{k} \right\rfloor$$

Here $\lfloor x \rfloor$ just means the whole-number part, so you divide and drop the remainder.

For example: multiples of 7 from 100 to 500. 500 ÷ 7 gives 71 and 99 ÷ 7 gives 14, so there are 71 − 14 = 57.

For "divisible by both 6 and 8", count the multiples of their LCM, 24. For "divisible by 6 or 8", add the two counts and subtract the multiples of 24 once, because they were counted twice.

Where it fails:

• Using a instead of a − 1. From 105 to 210 there are 16 multiples of 7, but 210 ÷ 7 − 105 ÷ 7 = 30 − 15 = 15 misses 105 itself.
• "Between 100 and 500" sometimes means both ends are left out. This only matters when an end is itself a multiple, so glance at the ends.
• "Both 6 and 8" means multiples of the LCM, 24, not of 6 × 8 = 48.

Sum of multiples: count × average of first and last

When to use it: "Find the sum of all multiples of k between a and b", or the sum of any list of numbers with a fixed gap.

Fast way:

$$\text{sum} = \text{count} \times \frac{\text{first} + \text{last}}{2}$$

For example: the multiples of 7 from 100 to 500 run from 105 to 497, and there are 57 of them. Sum = 57 × (105 + 497)/2 = 57 × 301 = 17,157.

Where it fails: first and last mean the first and last multiples, not the ends of the range. Using (100 + 500)/2 = 300 here gives 57 × 300 = 17,100, which is wrong and often sits in the options. The method also needs a fixed gap between terms. Three special cases worth knowing: $1 + 2 + \cdots + n = \frac{n(n+1)}{2}$, the first n odd numbers add up to $n^2$, and the first n even numbers add up to $n(n+1)$.

Factors and Zeros

Number of factors from the powers

When to use it: "How many factors (divisors) does N have?" and the odd or even versions.

Fast way: write N as a product of prime powers, $N = p^a \times q^b \times r^c$. The number of factors is $(a+1)(b+1)(c+1)$.

For example: 720 = $2^4 \times 3^2 \times 5$, so it has 5 × 3 × 2 = 30 factors. For the odd factors, leave out the 2s: 3 × 2 = 6. The even factors are the rest: 30 − 6 = 24.

Where it fails: the bases have to be primes. Writing 720 as $4^2 \times 45$ or as 8 × 9 × 10 and using those powers gives a wrong count. Also check what is being counted. "Factors" includes 1 and the number itself. "Factors other than the number" means subtract 1. A perfect square always has an odd number of factors, which is a quick check on your answer.

Zeros at the end of $n!$

When to use it: "How many zeros are there at the end of $250!$?" or at the end of 1 × 2 × 3 × … × n.

Fast way: divide n by 5 and drop the remainder. Divide that answer by 5 again, and keep going until you reach 0. Add up the answers.

For example: for $250!$: 250 ÷ 5 = 50, 50 ÷ 5 = 10, 10 ÷ 5 = 2, and 2 ÷ 5 = 0. Zeros = 50 + 10 + 2 = 62.

Where it fails:

• Stopping after the first division. 50 is the usual wrong option here, because it ignores the extra 5s inside 25, 50, 75, 125 and so on.
• This works for factorials because there are always more 2s than 5s. For any other product, count both and take the smaller. $2^4 \times 5^9$ has only 4 pairs of 2 and 5, so it ends in 4 zeros, not 9.

Perfect Squares

Quick check: can it be a perfect square?

When to use it: "Which of these is a perfect square?" or "Which of these cannot be a perfect square?"

Fast way: run three quick tests.

1. Last digit. A square never ends in 2, 3, 7 or 8, and never ends in an odd number of zeros.
2. Last two digits. If it ends in 5, it ends in 25. If it ends in 6, the tens digit is odd. If it ends in 1, 4 or 9, the tens digit is even.
3. Digit sum. Keep adding the digits until one digit is left. For a perfect square this is always 1, 4, 7 or 9.

For example: 59,048 ends in 8, so it is out. 63,246 ends in 46, and a square ending in 6 needs an odd tens digit, so it is out too. 31,684 passes all three tests (its digit sum is 22, then 4), and in fact 31,684 = $178^2$.

Where it fails: these tests can only reject a number. Passing them does not prove it is a square. 4,321 ends in 1 with an even tens digit, and its digit sum is 10, then 1. It still isn't a square, since $65^2 = 4225$ and $66^2 = 4356$. If two options survive, use the next method or square the likely root.

Square root of a 5-digit perfect square

When to use it: you know the number is a perfect square (the question says so, or every option is a whole number) and you want its root fast.

Fast way:

1. The last digit gives two choices for the root's last digit: 1 → 1 or 9, 4 → 2 or 8, 9 → 3 or 7, 6 → 4 or 6, 5 → 5, 0 → 0.
2. Cover the last two digits. Find the biggest square that fits into what is left. Its root, call it n, is the start of the answer.
3. If what is left is at least n × (n + 1), take the bigger choice. Otherwise take the smaller one.

For example: $\sqrt{54756}$. The last digit is 6, so the root ends in 4 or 6. Covering 56 leaves 547, and $23^2 = 529$ fits, so the root starts with 23. Now 23 × 24 = 552 is more than 547, so I take the smaller choice: 234.

Where it fails: it assumes the number really is a perfect square. Used on a non-square it still gives an answer, just a wrong one. The same steps work for 6-digit squares; what is left after covering then has 4 digits.

Smallest multiplier or divisor for a perfect square

When to use it: "By what least number must N be multiplied (or divided) to make it a perfect square?"

Fast way: factorise completely, pair up equal primes, and pick out the primes left with an odd power. Multiply or divide by their product.

For example: 1,188 = $2^2 \times 3^3 \times 11$. The odd powers belong to 3 and 11, so the answer is 3 × 11 = 33 both ways. 1,188 × 33 = 39,204 = $198^2$ and 1,188 ÷ 33 = 36 = $6^2$.

Where it fails: stopping at the first leftover prime. Many students spot the extra 3, answer 3, and forget the 11. For a perfect cube [Tier 2] every power must be a multiple of 3, so 1,188 needs one more 2 and two more 11s: multiply by 2 × $11^2$ = 242 to get $66^3$.

Remainder Shortcuts

Changing the divisor

When to use it: "A number leaves remainder 41 when divided by 56. What is the remainder when the same number is divided by 8?"

Fast way: if the new divisor is a factor of the old divisor, just divide the old remainder by the new divisor.

For example: 56 = 8 × 7, so 8 is a factor of 56. 41 ÷ 8 leaves 1, and that is the answer.

Where it fails: the new divisor must be a factor of the old one. Take 12, which does not divide 56. The numbers 41 and 97 both leave 41 when divided by 56, but they leave 5 and 1 when divided by 12. The answer isn't fixed, so the correct option is "cannot be determined". Going the other way, from a small divisor to a bigger one, can't be done like this either.

Negative remainders and $(d \pm 1)^n$

When to use it: a product or a big power is divided by d, and the numbers sit just above or just below multiples of d.

Fast way: replace each number by its distance from the nearest multiple of d, with a minus sign if it is below. For powers:

$$(d+1)^n \text{ leaves } 1 \qquad (d-1)^n \text{ leaves } (-1)^n$$

So $(d-1)^n$ leaves 1 when n is even and d − 1 when n is odd.

For example: 25 × 38 ÷ 13. 25 = 26 − 1 and 38 = 39 − 1, so the remainder is (−1) × (−1) = 1. For $17^{45}$ ÷ 18: 17 = 18 − 1, so the remainder is $(-1)^{45}$ = −1, which means 18 − 1 = 17.

A power can often be pushed into this shape. For $2^{50}$ ÷ 7: $2^3 = 8 = 7 + 1$, and $2^{50} = (2^3)^{16} \times 2^2$, so the remainder is 1 × 4 = 4.

Where it fails: a negative answer is not the final remainder. Add the divisor to it: −1 becomes d − 1 and −3 becomes d − 3. Choosing 1 for $17^{45}$ ÷ 18 is the standard mistake.

Unit Digits by Cyclicity

The 4-cycle table

When to use it: "Find the unit digit of $7^{154}$", or of a product, sum or difference of powers.

Fast way: only the last digit of the base matters. Divide the power by 4 and use the remainder to pick from the cycle. A remainder of 0 means the 4th entry.

To divide a huge power by 4, look only at its last two digits.

For example: $1237^{154}$. The base ends in 7. 154 ÷ 4 leaves 2, so I take the 2nd entry of 7, 9, 3, 1. The unit digit is 9.

Where it fails:

• A remainder of 0 means the 4th entry, not the 1st.
• For a difference, if the answer comes out negative, add 10. The unit digit of $7^{35} - 3^{35}$ is 3 − 7 = −4, so it is 6.
• In a product, anything ending in 5 times anything even ends in 0. Check for that before you reach for the table.

HCF Shortcuts

Same remainder: HCF of the differences

When to use it: "Find the greatest number that divides 653, 1,165 and 1,741 leaving the same remainder in each case."

Fast way: put the numbers in order, subtract neighbours, and find the HCF of the gaps. You never need the remainder itself.

For example: 1,165 − 653 = 512 and 1,741 − 1,165 = 576. HCF(512, 576) = 64, so the answer is 64. (The common remainder turns out to be 13.)

Where it fails: this is only for the same unknown remainder. If the question gives different remainders ("leaving 5, 7 and 9"), subtract each number's own remainder first and take the HCF of what is left. If it gives one stated remainder ("leaving 5 in each case"), subtract 5 from every number and take the HCF.

Ratio and HCF

When to use it: "Two numbers are in the ratio 4 : 7 and their HCF is 9", or "…and their LCM is 252".

Fast way: with the ratio a : b in lowest terms and HCF h, the numbers are ah and bh, and the LCM is a × b × h.

For example: ratio 4 : 7 and HCF 9 give the numbers 36 and 63, with LCM 4 × 7 × 9 = 252. Going the other way, ratio 3 : 5 and LCM 225 give h = 225 ÷ 15 = 15, so the numbers are 45 and 75.

Where it fails:

• The ratio must be in lowest terms first. 6 : 10 is really 3 : 5.
• For three numbers, the LCM is LCM(a, b, c) × h, not a × b × c × h. In the ratio 2 : 3 : 4 with HCF 5, the numbers are 10, 15 and 20, and the LCM is 12 × 5 = 60, not 24 × 5 = 120.

HCF × LCM = product

When to use it: you have two numbers' HCF and LCM and one of the numbers, and need the other.

Fast way: first number × second number = HCF × LCM.

For example: HCF 12, LCM 360, one number 72. The other is 12 × 360 ÷ 72 = 60.

Where it fails:

• It is true only for two numbers. For 4, 6 and 10, HCF × LCM = 2 × 60 = 120, but the product is 240.
• The HCF always divides the LCM. If a question says the HCF is 8 and the LCM is 60, no such pair exists, and you can strike out any option built on that.

LCM Shortcuts

Same remainder r: LCM + r

When to use it: "Find the least number which leaves remainder 4 when divided by 6, 9 and 15."

Fast way: LCM of the divisors + r. LCM(6, 9, 15) = 90, so the answer is 94.

With an extra condition: "…and is also divisible by 7." The number has the form 90k + 4. Try k = 1, 2, 3, …: 94, 184, 274, 364. The first one divisible by 7 is 364.

Where it fails:

• r has to be smaller than every divisor, or the question makes no sense.
• Strictly, 4 itself leaves remainder 4 on division by 6, 9 and 15. SSC means a number bigger than the divisors, which is why the expected answer is LCM + r.
• With an extra condition, test k one at a time and stop at the first one that works. Don't assume k = 1.

Constant gap: LCM − gap

When to use it: "Find the least number which leaves remainders 3, 5 and 9 when divided by 5, 7 and 11." Check divisor − remainder each time: 5 − 3 = 2, 7 − 5 = 2, 11 − 9 = 2. The gap is the same.

Fast way: LCM − gap.

For example: LCM(5, 7, 11) = 385, so the answer is 385 − 2 = 383.

Where it fails: only when every gap is equal. If the gaps differ, this shortcut doesn't apply, and in the exam the quickest way is usually to test the options against each divisor.

HCF or LCM? Decide from what is being found

For example: bells ringing every 12, 15 and 20 minutes ring together again after LCM = 60 minutes. A floor 4.2 m × 3.6 m takes square tiles of side HCF(420, 360) = 60 cm, and (420 ÷ 60) × (360 ÷ 60) = 7 × 6 = 42 tiles.

Where it fails: "least number of tiles" has the word least in it, but the tile size is still an HCF. Decide by what is being found (a size that fits into things, or a time or length that everything fits into), not by the words least or greatest. Change every measurement to the same unit, usually cm, before you start.

Option Elimination

SSC options are often close together, but they rarely survive all of these checks. Use them when the direct method is slow, or to confirm an answer in two seconds.

Last digit

When to use it: products, squares, powers and any long multiplication.

Fast way: work out only the unit digit of the answer and strike out every option that doesn't match.

For example: 347 × 286. 7 × 6 = 42, so the answer ends in 2. Of 99,242, 99,244, 99,246 and 99,248, only 99,242 survives.

Where it fails: when two options share the last digit. Then add a second check.

Digit sum and divisibility

When to use it: two options survive the last-digit check, or the question is about HCF and LCM.

Fast way: the digit sum of a product follows the digit sums of its parts. For 678 × 243: 678 → 21 → 3 and 243 → 9, and 3 × 9 = 27 → 9, so the answer is a multiple of 9. Between 1,64,754 and 1,65,754 (both end in 4), only 1,64,754 has a digit sum of 27. In HCF and LCM questions, the HCF must divide every given number and the LCM must be a multiple of every one, so one quick division often clears two options.

Where it fails: digit sums can't separate options that differ by a multiple of 9. If that happens, go back to the last digit or the size.

Size

When to use it: roots, large products, and HCF or LCM answers.

Fast way: estimate. The HCF is never bigger than the smallest number, and the LCM is never smaller than the biggest. The square root of a 5-digit number is a 3-digit number from 100 to 316.

For example: $\sqrt{60025}$ with options 235, 245, 255 and 265. All end in 5, so the last digit is no help. But $240^2 = 57600$ and $250^2 = 62500$, so the root lies between 240 and 250: 245.

Where it fails: estimation only separates options that are far apart. When the options differ by 1 or 2 (remainders, missing digits), do the actual working.

Solved Examples

Each answer shows the fast route, then a time check comparing it with the long way.

Question 1

Write $1.2\overline{45}$ as a fraction in lowest terms.

Answer. I keep the 1 aside and work on $0.2\overline{45}$. On top: 245 − 2 = 243. Two digits repeat and one doesn't, so the bottom is 990. That gives $\frac{243}{990} = \frac{27}{110}$. Adding the 1 back: $1 + \frac{27}{110} = \mathbf{\frac{137}{110}}$.

Time check: about 15 seconds. Writing x = 1.2454…, multiplying by 10 and by 1000 and subtracting takes close to a minute.

Question 2

If the six-digit number 3x462y is divisible by 72, find x + y.

Answer. 72 = 8 × 9, so I do the 8 test first. The last three digits 62y must be a multiple of 8. 624 = 8 × 78 works and no other digit does, so y = 4. Now the 9 test: 3 + x + 4 + 6 + 2 + 4 = 19 + x, and that must be 27, so x = 8. So x + y = 12, and the number is 3,84,624.

Time check: about 20 seconds. Trying digit pairs one by one could mean dozens of divisions.

Question 3

If 86x031 is divisible by 11, find x.

Answer. Odd places from the right: 1 + 0 + 6 = 7. Even places: 3 + x + 8 = 11 + x. The difference is 7 − (11 + x) = −4 − x. It can't be 0, because x would be −4. So it must be −11, which gives x = 7. Check: 8,67,031 = 11 × 78,821.

Time check: 15 seconds. The trap is stopping at "difference 0 gives −4, so no answer".

Question 4

What is the least number that must be added to 50,000 to make it exactly divisible by 47?

Answer. 47 × 1,000 = 47,000 leaves 3,000, and 47 × 63 = 2,961, so 50,000 = 47 × 1,063 + 39. The remainder is 39, so I add 47 − 39 = 8. Don't mark 39: that is what you would subtract, and SSC usually puts it among the choices.

Time check: one division, about 25 seconds. Testing 50,001, 50,002 and so on takes much longer.

Question 5

How many numbers from 150 to 750 (both included) are divisible by 4 or 6?

Answer. Multiples of 4: 750 ÷ 4 gives 187 and 149 ÷ 4 gives 37, so 150. Multiples of 6: 125 − 24 = 101. Numbers divisible by both are multiples of 12: 62 − 12 = 50. They got counted twice, so I take them off once: 150 + 101 − 50 = 201.

Time check: about 30 seconds. Listing the multiples is impossible in exam time.

Question 6

Find the sum of all the multiples of 17 between 200 and 800.

Answer. The first multiple is 204 (17 × 12) and the last is 799 (17 × 47). The count is 47 − 11 = 36. Sum = 36 × (204 + 799)/2 = 18 × 1003 = 18,054. I use 204 and 799, not 200 and 800, because the range ends are not multiples of 17.

Time check: about 30 seconds. Adding 36 numbers one by one is not an option.

Question 7

How many factors does 1,800 have, and how many of them are odd?

Answer. 1,800 = 18 × 100 = $2^3 \times 3^2 \times 5^2$. Factors: 4 × 3 × 3 = 36. For the odd ones I leave out the 2s: 3 × 3 = 9.

Time check: about 20 seconds. Listing 36 factors in pairs takes two minutes, and one usually gets missed.

Question 8

How many zeros are there at the end of $380!$?

Answer. 380 ÷ 5 = 76, 76 ÷ 5 = 15 (dropping the remainder), 15 ÷ 5 = 3, and 3 ÷ 5 = 0. Total: 76 + 15 + 3 = 94.

Time check: 10 seconds. There is no other way that fits in an exam.

Question 9

Find $\sqrt{75076}$.

Answer. The last digit is 6, so the root ends in 4 or 6. Covering the last two digits leaves 750, and $27^2 = 729$ fits, so the root starts with 27. 27 × 28 = 756 is more than 750, so I take the smaller choice: 274.

Time check: about 15 seconds. Prime factorising 75,076 takes over a minute.

Question 10

What is the smallest number by which 7,425 must be divided so that the result is a perfect square?

Answer. 7,425 = 25 × 297 = 25 × 27 × 11 = $3^3 \times 5^2 \times 11$. The 3 and the 11 have odd powers, so I divide by 3 × 11 = 33. The result is 225 = $15^2$. Dividing by 3 alone is the trap, because it leaves the 11 unpaired.

Time check: 20 seconds. Dividing by each option and then checking for a square takes about a minute.

Question 11

Find the remainder when $5^{63}$ is divided by 31.

Answer. $5^3 = 125 = 4 \times 31 + 1$, so $5^3$ leaves 1. And $5^{63} = (5^3)^{21}$, so the remainder is $1^{21}$ = 1.

Time check: 15 seconds. Writing out the remainders of 5, 25, 125, … until a pattern appears takes much longer and invites mistakes.

Question 12

Find the unit digit of $2^{38} \times 3^{57} \times 7^{43}$.

Answer. 38 ÷ 4 leaves 2, so $2^{38}$ ends in 4. 57 ÷ 4 leaves 1, so $3^{57}$ ends in 3. 43 ÷ 4 leaves 3, so $7^{43}$ ends in 3. Then 4 × 3 × 3 = 36, so the unit digit is 6.

Time check: about 20 seconds. Without the table there is no sensible way to do this.

Question 13

Find the greatest number that divides 2,128, 2,633 and 3,441 leaving the same remainder in each case.

Answer. I don't need the remainder at all. The gaps are 2,633 − 2,128 = 505 and 3,441 − 2,633 = 808. 505 = 5 × 101 and 808 = 8 × 101, so their HCF is 101. (Each number leaves remainder 7.)

Time check: 20 seconds. Guessing divisors and testing them would never finish.

Question 14

Find the least number which leaves remainder 5 when divided by 12, 16 and 18, and is exactly divisible by 7.

Answer. LCM(12, 16, 18) = 144, so the number is 144k + 5. Instead of dividing each candidate by 7, I use remainders: 144 leaves 4 when divided by 7, so I need 4k + 5 to be a multiple of 7. k = 4 gives 21. So the number is 144 × 4 + 5 = 581, and 581 = 7 × 83.

Time check: about 30 seconds. Checking 149, 293 and 437 by long division first would double that.

Question 15

$\sqrt{93025}$ = ? (a) 295 (b) 305 (c) 315 (d) 325

Answer. All four options end in 5, so the last digit is no help. I check size instead: $300^2 = 90000$ and $310^2 = 96100$. 93,025 lies between them, so the root is between 300 and 310. Only (b) 305 fits.

Time check: 10 seconds. Squaring options one by one takes about a minute.