Remainders and Unit Digits

Dividend, Divisor, Quotient and Remainder

Every remainder question comes back to one line:

$$\text{Dividend} = \text{Divisor} \times \text{Quotient} + \text{Remainder}$$

When 100 is divided by 7, the quotient is 14 and the remainder is 2, because 7 × 14 + 2 = 100.

Two rules hold every single time:

• The remainder is always less than the divisor. If you divide by 7 and get a remainder of 9, you stopped too early. So if a question says the remainder is 39, the divisor has to be at least 40.
• The final remainder is never negative. It can be 0, which just means the number is divisible. Negative remainders are a handy working trick (coming up next), but the answer you mark always lies between 0 and one less than the divisor.

Finding the dividend from relations

SSC often hides the divisor and the quotient inside "times" statements. Say the divisor is 14 times the quotient and 4 times the remainder, and the remainder is 35. Start from the one quantity you actually know and walk outwards:

• Divisor = 4 × remainder = 4 × 35 = 140
• Quotient = 140 ÷ 14 = 10
• Dividend = 140 × 10 + 35 = 1435

Check: 1435 ÷ 140 gives 10 and leaves 35, and 35 is less than 140, so it holds up.

Wrong-divisor questions

Someone divides by the wrong number. The dividend never changes, so rebuild it from the wrong division and then divide it properly.

A boy had to divide a number by 15 but divided it by 35 instead, getting quotient 27 and remainder 22. The number is 35 × 27 + 22 = 967. The correct division is 967 ÷ 15 = 64, remainder 7.

When the wrong division comes out exact, it is even quicker. A student used 72 as the divisor instead of 27 and got 15. The number is 72 × 15 = 1080, so the right answer is 1080 ÷ 27 = 40.

Successive division

"A number is divided by 4, and the quotient is then divided by 5" is called successive division. Work from the last step backwards. Say the remainders are 1 and 4. Call the first quotient $q$. Since $q$ leaves 4 when divided by 5, $q = 5k + 4$. The number is

$$4q + 1 = 4(5k + 4) + 1 = 20k + 17$$

The smallest such number is 17 (take $k = 0$), and every such number leaves remainder 17 when divided by 4 × 5 = 20.

Remainders of Sums, Differences and Products

This is the rule that saves the most time in the whole topic: to find the remainder of a sum, a difference or a product, replace each number by its remainder first, work with those small numbers, and then take the remainder once more.

Take 47 and 58, divided by 9. 47 leaves 2 and 58 leaves 4.

• Sum: 2 + 4 = 6, so 47 + 58 = 105 leaves 6. (Check: 9 × 11 + 6 = 105.)
• Product: 2 × 4 = 8, so 47 × 58 = 2726 leaves 8. (Check: 9 × 302 + 8 = 2726.)
• Difference: 58 − 47 leaves 4 − 2 = 2.

If a difference of remainders comes out negative, add the divisor. For 47 − 31: 47 leaves 2 and 31 leaves 4, and 2 − 4 = −2. Adding 9 gives 7, and sure enough 16 leaves 7 when divided by 9.

Negative remainders

A number just below a multiple of the divisor can be given a negative remainder. 1235 is one less than 1236 = 12 × 103, so for division by 12 we can call its remainder −1 instead of 11. Both are correct. −1 is just much easier to multiply.

Find the remainder when 1235 × 1237 × 1239 is divided by 12. The remainders are 11 (or −1), 1 and 3.

• With ordinary remainders: 11 × 1 × 3 = 33, and 33 leaves 9.
• With a negative remainder: (−1) × 1 × 3 = −3, and −3 + 12 = 9.

Same answer, smaller numbers. Use whichever remainder is smaller in size, and if the final answer is negative, add the divisor to bring it back between 0 and the divisor. This pays off most when the numbers sit right next to a multiple, like 99 or 1001 next to 100 or 1000.

Finding the divisor from remainders

Two numbers leave remainders 17 and 23 when divided by the same divisor, and their sum leaves remainder 9. If nothing had spilled over, the sum would leave 17 + 23 = 40. It leaves only 9, so the extra 40 − 9 = 31 must be one full divisor. The divisor is 31, and it is bigger than both remainders, so it fits.

Changing the Divisor, and Remainders of Expressions

Changing the divisor

A number leaves remainder 37 when divided by 65. What does it leave when divided by 13?

Write the number as $65k + 37$. Since 65 = 13 × 5, the $65k$ part is a multiple of 13 and leaves nothing. Only the 37 matters, and 37 = 13 × 2 + 11, so the remainder is 11.

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

This works only when the new divisor goes exactly into the old one. Ask the same number's remainder by 15 instead. 65 is not a multiple of 15, so the $65k$ part no longer disappears, and different values of $k$ give different answers: 37 leaves 7, 102 leaves 12 and 167 leaves 2. There is no single answer.

The reverse doesn't work either. If a number leaves 11 when divided by 13, its remainder by 65 could be 11, 24, 37, 50 or 63, because all five of these leave 11 when divided by 13. A remainder by a small divisor can never pin down the remainder by a bigger one. If an SSC question seems to ask for that, check which divisor is the factor of which.

Remainder of an expression

If you know what $n$ leaves, you know what any expression in $n$ leaves. Put the remainder in place of $n$ and work it out.

• $n$ leaves 4 when divided by 9. For $n^2 + 2n + 7$, put $n = 4$: 16 + 8 + 7 = 31, and 31 leaves 4 when divided by 9. So the remainder is 4.
• $m$ leaves 7 when divided by 9. Then $5m$ leaves the same as 5 × 7 = 35, which is 8.
• A number leaves 8 when divided by 15. Its square leaves the same as 8 × 8 = 64, which is 4.

Why does this work? Write $n = 9k + 4$. When you expand $n^2 + 2n + 7$, every term with $k$ in it is a multiple of 9, so only the part made from the 4 survives.

[SSC Tip] The fastest check is to pick the smallest number that fits. If $n$ leaves 4 when divided by 9, take $n = 4$ or $n = 13$ and just calculate. It takes ten seconds and catches arithmetic slips.

Remainders of Large Powers: the Divisor ± 1 Trick

Powers like $19^{200}$ look impossible, but the remainder of a power depends only on the remainder of the base. So the whole game is to make the base leave 1 or −1.

Base is one more than the divisor. Then the base leaves 1, and 1 to any power is still 1. So $(d + 1)^n$ divided by $d$ always leaves 1. For example, $37^{94}$ divided by 36 leaves 1, and so does $37^{94}$ divided by 9, because 37 = 36 + 1 and 36 is a multiple of 9.

Base is one less than the divisor. Then the base leaves −1. And $(-1)^n$ is 1 when $n$ is even and −1 when $n$ is odd. So:

$17^{150}$ divided by 18: 17 is −1 and 150 is even, so the remainder is 1. $23^{75}$ divided by 24: 75 is odd, so we get −1, which is 23.

With a number added. Handle the power, then add.

• $341^{1205} + 13$ divided by 342: 341 is −1 and the power is odd, so −1 + 13 = 12.
• $12^{12} + 7$ divided by 11: 12 leaves 1, so 1 + 7 = 8.
• $11^{25} + 2$ divided by 4: 11 leaves 3, which is −1, the power is odd, so −1 + 2 = 1.

If the total goes past the divisor, take the remainder again.

The $x^x + x$ type. $49^{49} + 49$ divided by 50: 49 is −1, so $49^{49}$ leaves −1 (odd power) and the lone 49 leaves −1 as well. Total −2, and −2 + 50 = 48. In general, for odd $x$, $x^x + x$ divided by $x + 1$ leaves $x - 1$. For even $x$ the power gives +1 instead, so the total is 1 + (−1) = 0: $48^{48} + 48$ is exactly divisible by 49.

When the base is not next to the divisor, group the power so that it is. For $2^{100}$ divided by 7, notice that $2^3 = 8$ is one more than 7. So

$$2^{100} = 2^{99} \times 2 = (2^3)^{33} \times 2$$

$(2^3)^{33}$ leaves $1^{33} = 1$, so the answer is 1 × 2 = 2.

[SSC Tip] Keep these pairs in your head: $2^3 = 8$ sits next to 7 and 9, $2^4 = 16$ next to 15 and 17, $2^5 = 32$ next to 31 and 33, $3^3 = 27$ next to 26 and 28, $3^4 = 81$ next to 80 and 82, and $5^2 = 25$ next to 24 and 26. Power questions in SSC papers are often built on one of them.

Cycles of Remainders, and Fermat's Little Theorem

When grouping isn't obvious, write down the remainders of the first few powers. They always fall into a repeating cycle.

The cycle 2, 4, 1 has length 3. For $2^{65}$ divided by 7, divide the power by the cycle length: 65 ÷ 3 leaves 2, so take the 2nd value in the cycle, which is 4. If the power divides exactly by the cycle length, take the last value of the cycle.

A few more cycles worth knowing:

• Powers of 3 divided by 7: 3, 2, 6, 4, 5, 1, a cycle of length 6.
• Powers of 2 divided by 5: 2, 4, 3, 1, a cycle of length 4.
• Powers of 3 divided by 8: 3, 1, so odd powers leave 3 and even powers leave 1.

The method in three steps:

1. Write the remainders of the base, its square, its cube and so on, until you reach 1.
2. The number of steps it took to reach 1 is the cycle length.
3. Divide the given power by the cycle length. The remainder of that division tells you the position in the cycle (0 means the last one).

Reaching 1 is the important part, because once a power leaves 1, everything after it repeats from the start.

Fermat's little theorem [Tier 2]

If $p$ is a prime and $a$ is not a multiple of $p$, then

$$a^{p - 1} \text{ leaves remainder 1 when divided by } p$$

So for division by 7, any $a^6$ leaves 1. For 11, any $a^{10}$ leaves 1. For 13, any $a^{12}$ leaves 1.

$3^{100}$ divided by 7: $3^6$ leaves 1, and 100 = 6 × 16 + 4, so $3^{100} = (3^6)^{16} \times 3^4$. That leaves the same as $3^4 = 81$, and 81 = 7 × 11 + 4, so the remainder is 4.

$4^{50}$ divided by 13: $4^{12}$ leaves 1, and 50 = 48 + 2, so the answer is the remainder of $4^2 = 16$, which is 3.

The real cycle can be shorter than $p - 1$ (for powers of 2 divided by 7 it is 3, a factor of 6), but $a^{p - 1}$ always lands on 1, so the theorem never misleads you. It only works for prime divisors, though. For divisors like 8, 9 or 12, use the cycle method or the ±1 trick.

Quick Remainders by 9, 11, 4, 8, 16, 5 and 25

Some divisors let you read the remainder straight off the digits.

By 9 (and 3). A number leaves the same remainder as its digit sum. For 7,46,385 the digit sum is 7 + 4 + 6 + 3 + 8 + 5 = 33, and 33 leaves 6 when divided by 9, so 746385 leaves 6. For 3, use the same digit sum: 33 leaves 0, so 746385 is divisible by 3. This also handles repeated digits: 555…5 with thirty 5s has digit sum 150, which leaves 6.

By 11. Start at the units digit. Add the digits in the odd places (1st, 3rd, 5th, … from the right) and subtract the digits in the even places. The remainder of that result by 11 is the answer, and if it is negative, add 11 until it isn't.

• 8,73,926: odd places 6 + 9 + 7 = 22, even places 2 + 3 + 8 = 13, and 22 − 13 = 9. Remainder 9.
• 9182: odd places 2 + 1 = 3, even places 8 + 9 = 17, and 3 − 17 = −14. Add 22 to get 8. Remainder 8.

Always start from the right. Starting from the left gives the wrong sign whenever the number has an even number of digits.

By 2, 4, 8, 16. 10 is divisible by 2, 100 by 4, 1000 by 8 and 10000 by 16, so everything before the last 1, 2, 3 or 4 digits disappears. 8,73,926 divided by 4: look at 26, which leaves 2. Divided by 8: look at 926 = 8 × 115 + 6, so it leaves 6.

By 5, 10 and 25. Use the last digit for 5 and 10, and the last two digits for 25. 87,643 divided by 25: 43 leaves 18.

Least number to add or subtract

Once you know the remainder $r$ when a number is divided by $d$:

• the least number to subtract to get a multiple of $d$ is $r$,
• the least number to add is $d - r$ (when $r$ is not 0).

23,659 divided by 17 leaves 12. Subtract 12 to get 23,647, or add 17 − 12 = 5 to get 23,664. Both are multiples of 17. The versions with several divisors and with largest and smallest n-digit numbers are in the divisibility section.

Unit Digits and Cyclicity

The unit digit of a product depends only on the unit digits of the numbers being multiplied. So $158^{175}$ ends in the same digit as $8^{175}$, and you never need the 15.

The unit digits of powers repeat:

Every cycle length (1, 2 or 4) fits into 4, so one method covers everything:

1. Take the last digit of the base.
2. Divide the power by 4 and note the remainder $r$. Only the last two digits of the power matter for this.
3. Call the last digit $u$. If $r$ is 1, 2 or 3, the answer is the unit digit of $u^r$. If $r$ is 0, use $u^4$.

$158^{175}$: last digit 8, and 175 ÷ 4 leaves 3. $8^3 = 512$, so the unit digit is 2.

$2^{100}$: 100 ÷ 4 leaves 0, so use $2^4 = 16$. The unit digit is 6.

For 4 and 9 you only need odd or even: odd powers of 4 end in 4 and even powers in 6, odd powers of 9 end in 9 and even powers in 1. Numbers ending in 0, 1, 5 and 6 always end in the same digit, whatever the power.

Products. Work out each unit digit, multiply, keep the last digit. For 123 × 456 × 789: 3 × 6 × 9 = 162, so it ends in 2. For $7^{35} \times 3^{71} \times 11^{55}$: 35 and 71 both leave 3 when divided by 4, so $7^{35}$ ends in 3 and $3^{71}$ ends in 7, and $11^{55}$ ends in 1. 3 × 7 × 1 = 21, so the product ends in 1.

Sums and differences. Add or subtract the unit digits. $2^{51} + 3^{62}$: 51 leaves 3, so $2^{51}$ ends in 8, and 62 leaves 2, so $3^{62}$ ends in 9. 8 + 9 = 17, so the sum ends in 7.

For differences, borrow when you need to. $7^{95} - 3^{58}$: $7^{95}$ ends in 3 and $3^{58}$ ends in 9. 3 − 9 can't be done, so borrow: 13 − 9 = 4. This works because $7^{95}$ is the bigger number.

Unit Digits of Factorial Sums and Long Products, and Last Two Digits

Factorial sums

$n! = 1 \times 2 \times 3 \times \dots \times n$. From $5! = 120$ onwards, every factorial has both a 2 and a 5 in it, so it ends in 0. In a sum like $1! + 2! + 3! + \dots + 100!$ only the first four terms touch the unit digit:

$$1 + 2 + 6 + 24 = 33$$

So the unit digit is 3, and it is 3 for any such sum that starts at $1!$ and goes past $4!$. Change the starting point and the answer changes too. A sum starting from $3!$ ends like 6 + 24 = 30, in 0. A product of factorials that includes $5!$ ends in 0.

Long products

• The product of the first 20 natural numbers has a 2 and a 5, so it ends in 0.
• The product of all odd numbers from 1 to 99 is odd and has a 5 in it, so it ends in 5.
• The product of all primes below 50 has 2 and 5 in it, so it ends in 0.
• 21 × 22 × 23 × 24 ends like 1 × 2 × 3 × 4 = 24, so in 4.

Before you multiply anything, scan for a 5 (or a number ending in 5) and an even number. If both are there, the answer is 0 and you are done.

Last two digits [Tier 2]

Two methods cover most Tier 2 questions.

Bases ending in 1. For a base like 31 or 71, the last digit of any power is 1, and the tens digit is the unit digit of (tens digit of the base) × (power). For $31^{46}$: 3 × 46 = 138, so the tens digit is 8 and the last two digits are 81. This works because $(10a + 1)^n = 1 + 10an + \text{(multiples of 100)}$.

Powers of 7. $7^4 = 2401$ ends in 01, so every $7^{4k}$ ends in 01. For $7^{42}$, write $7^{42} = 7^{40} \times 7^2$. The first part ends in 01, so the answer ends like $7^2 = 49$. The same idea works for 3: $3^4 = 81$ ends in 1, so $3^{20} = 81^5$, and the base-ending-in-1 rule gives tens digit 8 × 5 = 40, which is 0. So $3^{20}$ ends in 01.

For anything else, multiply step by step and keep only the last two digits each time. It is slower, but it never goes wrong.

Solved Examples

Question 1

In a division sum, the divisor is 9 times the quotient and 3 times the remainder. If the remainder is 24, what is the dividend?

Answer. The remainder is the only number I actually know, so I start there. Divisor = 3 × 24 = 72. The divisor is also 9 times the quotient, so the quotient is 72 ÷ 9 = 8. Dividend = 72 × 8 + 24 = 576 + 24 = 600.

Question 2

In a division sum, the divisor is 25 times the quotient and 5 times the remainder. If the quotient is 16, find the dividend.

Answer. This time the quotient is given. Divisor = 25 × 16 = 400. Remainder = 400 ÷ 5 = 80. Dividend = 400 × 16 + 80 = 6400 + 80 = 6480. Quick check: 80 is less than 400, so the remainder is allowed.

Question 3

A student had to divide a number by 16. By mistake he divided it by 36 and got 38 as the quotient and 29 as the remainder. What are the correct quotient and remainder?

Answer. The number itself doesn't change, so I rebuild it: 36 × 38 + 29 = 1368 + 29 = 1397. Now the right division: 16 × 87 = 1392, and 1397 − 1392 = 5. So the correct quotient is 87 and the remainder is 5.

Question 4

A number is divided successively by 3 and 7, leaving remainders 2 and 5. What remainder does it leave when divided by 21?

Answer. I work backwards. The first quotient leaves 5 when divided by 7, so it is $7k + 5$. The number is $3(7k + 5) + 2 = 21k + 17$. Whatever $k$ is, $21k$ divides exactly by 21, so the remainder is 17. The smallest such number is 17 itself: 17 ÷ 3 = 5 remainder 2, and 5 ÷ 7 = 0 remainder 5.

Question 5

Find the remainder when 523 × 527 × 531 is divided by 25.

Answer. I don't multiply. 523 = 525 − 2, and 525 is a multiple of 25, so 523 leaves −2. 527 leaves 2 and 531 leaves 6. Now (−2) × 2 × 6 = −24. A remainder can't be negative, so I add 25: −24 + 25 = 1.

Question 6

Two numbers leave remainders 19 and 26 when divided by the same divisor. Their sum leaves remainder 7 when divided by that divisor. Find the divisor.

Answer. If nothing had spilled over, the sum would leave 19 + 26 = 45. It leaves only 7, so one full divisor got used up: 45 − 7 = 38. And 38 is bigger than 26, so both remainders make sense.

Question 7

A number leaves remainder 47 when divided by 342. What remainder does it leave when divided by 18?

Answer. First I check that 18 goes into 342: 342 = 18 × 19, so it does. That means the 342 part disappears and only the 47 matters. 47 = 18 × 2 + 11, so the remainder is 11.

Question 8

When $n$ is divided by 8, the remainder is 5. What is the remainder when $n^2 + 4n + 3$ is divided by 8?

Answer. I put 5 in place of $n$: 25 + 20 + 3 = 48. And 48 = 8 × 6, so the remainder is 0. To be safe I try $n = 13$ as well: 169 + 52 + 3 = 224 = 8 × 28. Still 0.

Question 9

A number leaves remainder 10 when divided by 13. What remainder does its square leave when divided by 13?

Answer. The square leaves the same as 10 × 10 = 100. 13 × 7 = 91, and 100 − 91 = 9. A quicker way: 10 is −3 for 13, and (−3) × (−3) = 9.

Question 10

Find the remainder when $25^{180}$ is divided by 26.

Answer. 25 is one less than 26, so it leaves −1. The power 180 is even, so $(-1)^{180} = 1$. The remainder is 1.

Question 11

Find the remainder when $313^{2017} + 21$ is divided by 314.

Answer. 313 is −1 for 314. The power 2017 is odd, so $313^{2017}$ leaves −1. Then −1 + 21 = 20.

Question 12

What is the remainder when $9^9 + 5$ is divided by 8?

Answer. 9 is one more than 8, so it leaves 1, and $1^9$ is still 1. So the remainder is 1 + 5 = 6.

Question 13

Find the remainder when $59^{59} + 59$ is divided by 60.

Answer. 59 is −1 for 60. The power is odd, so $59^{59}$ leaves −1, and the 59 on its own leaves −1 too. That is −2 in total, and −2 + 60 = 58. This always happens: for odd $x$, $x^x + x$ divided by $x + 1$ leaves $x - 1$.

Question 14

Find the remainder when $2^{31}$ is divided by 15.

Answer. 2 is nowhere near 15, but $2^4 = 16$ is one more than 15, so it leaves 1. I split the power: 31 = 4 × 7 + 3, so $2^{31} = (2^4)^7 \times 2^3$. The first part leaves $1^7 = 1$, so the remainder is 1 × 8 = 8.

Question 15 [Tier 2]

Find the remainder when $5^{62}$ is divided by 13.

Answer. 13 is prime and 5 is not a multiple of 13, so by Fermat's little theorem $5^{12}$ leaves 1. 62 = 12 × 5 + 2, so $5^{62} = (5^{12})^5 \times 5^2$, which leaves 1 × 25 = 25. And 25 = 13 + 12, so the remainder is 12.

Another route: $5^2 = 25$ is one less than 26, so it is −1 for 13. Then $5^{62} = (5^2)^{31}$ leaves $(-1)^{31} = -1$, which is 12. Same answer.

Question 16

Without dividing, find the remainders when 5,29,174 is divided by 9, by 11 and by 8.

Answer. By 9: the digit sum is 5 + 2 + 9 + 1 + 7 + 4 = 28, and 28 leaves 1.

By 11: from the right, the odd places hold 4, 1, 2 (sum 7) and the even places hold 7, 9, 5 (sum 21). So 7 − 21 = −14, and adding 22 gives 8.

By 8: the last three digits are 174 = 8 × 21 + 6, so it leaves 6.

The remainders are 1, 8 and 6.

Question 17

What is the remainder when 777…7 (twenty-five 7s) is divided by 9?

Answer. A number leaves the same remainder as its digit sum when divided by 9. The digit sum is 25 × 7 = 175. Its own digit sum is 1 + 7 + 5 = 13, which leaves 4. The remainder is 4.

Question 18

What is the least number that must be subtracted from 9876 to make it divisible by 37? What is the least number that must be added?

Answer. 37 × 266 = 9842, and 9876 − 9842 = 34, so 9876 leaves 34. Taking away the remainder gives a multiple, so I subtract 34. To reach the next multiple I add 37 − 34 = 3, and 9879 = 37 × 267 confirms it.

Question 19

Find the unit digit of $263^{157}$.

Answer. Only the last digit 3 matters. Powers of 3 end in 3, 9, 7, 1 and then repeat, a cycle of 4. 157 ÷ 4 leaves 1, so it is the first one in the cycle. The unit digit is 3.

Question 20

Find the unit digit of $3^{65} \times 6^{59} \times 7^{71}$, and of $3^{65} + 6^{59} + 7^{71}$.

Answer. 65 ÷ 4 leaves 1, so $3^{65}$ ends in 3. Every power of 6 ends in 6. 71 ÷ 4 leaves 3, so $7^{71}$ ends like $7^3 = 343$, in 3.

Product: 3 × 6 × 3 = 54, so it ends in 4. Sum: 3 + 6 + 3 = 12, so it ends in 2.

Question 21

What is the unit digit of $4! + 5! + 6! + \dots + 40!$?

Answer. Every factorial from $5!$ onwards has a 2 and a 5 in it, so it ends in 0. Only $4! = 24$ adds anything to the last digit. The unit digit is 4. I have to be careful here, because the answer 3 that everyone remembers is for sums that start at $1!$.

Question 22 [Tier 2]

Find the last two digits of $21^{38}$.

Answer. The base ends in 1, so the last digit is 1. The tens digit is the unit digit of 2 × 38 = 76, which is 6. So the last two digits are 61.