Number System is known for intriguing conceptual problems that test the best brains. It requires combination of pure mathematical knowledge and logic skills. Keeping in mind the vast syllabus of number system in CAT, we will cover only the concepts and tricks which are less known yet important.

### 1. Zeroes in *n!*

Greatest power of prime number, *p*, which divides *n!* is *[n!/p] + [n!/p ^{2}] + [n!/p^{3}] + . . .* , where [ . ] is greatest integer function. Note that

**this is only valid when**. For example, greatest power of 7 that divides 120! is [120/7] + [120/49] + [120/343] = 17 + 2 = 19.

*p*is prime numberNow to find number of zeroes at the end of number *n!*, we need to find greatest power of 10 that divides *n!*. But 10 is not a prime number, it has factors 2 and 10 (apart from obvious 1 and 10).

Lets assume *n! = 2 ^{a} *×

*5*×

^{b}*c*, and we can definitely say that a > b. So we need to find greatest power of 5 that divides

*n!*and that will be same as the greatest power of 10 that divides

*n!*. That would give us-

*[n/5] + [n/25] + [n/125] + [n/625] + . . .*

For example, number of trailing zeroes in 120! would be [120/5] + [120/25] + [120/125] = 24 + 4 + 0 = 28. Simple, isn’t it?

### 2. HCF Shortcut

Let’s say you want to find HCF of numbers *a, b, c* and *d* (*wlog a ≤ b ≤ c ≤ d*). Rather then going by normal approach, its much faster if you find HCF of *a*, *b – a*, *c – b*, *d – c*. Remember that HCF of *a* and *b* is same as HCF of *a* and *b – a*.

Let’s take an example- Find HCF of 144, 630 and 756.

630 – 144 = 486, 756 – 630 = 126. So we need to find HCF of 144, 486 and 126. Doing it again, 144 – 126 = 18, 486 – 144 = 342. That gives HCF of 18, 126, 342.

It is now easy to see that all numbers are divisible by 9 and 2, hence 18 is HCF of 144, 630 and 756.

Note: we could have also done 630 – 4 × 144 = 54.

### 3. LCM Shortcut

By using concept of co-primes, we can bypass the need to find prime factors of all the numbers for which we are trying to find LCM.

Suppose we want to find LCM of 9, 10, 12, 15.

**Step 1**: If you see a set of 2 or more co-primes, take multiplication of them, i.e. 9 × 10.**Step 2**: For each of the other numbers, consider what factors of them have already be taken in the answer and what part remains outside the answer.

Factors of 12 are 2 × 2 × 3. One 2 and one 3 are already in 9 × 10. So the updated answer becomes 2 × 9 × 10.

Factors of 15 are 3 × 5. Both 3 and 5 are already considered in the answer. So the answer remains 2 × 9 × 10.

Hence the LCM is 180.

### 4. Divisibility by 7, 11 and 13

A number is divisible by 7, 11 or 13 if and only if the difference of the number of its thousands and the remainder of its division by 1000 is divisible by 7, 11 or 13 respectively. This works because 7 × 11 × 13 = 1001.

For example, 473312 is divisible by 7 since 473 – 312 = 161 is divisible by 7.

Note: You can even find remainder by this method.

### 5. Unit’s digit

We need to understand concept of cyclicity, mainly about the unit digit of a number and its repetitive pattern on being divided by a certain number. The numbers can be divided into three categories-

- Digits 0, 1, 5 & 6: All the powers of these numbers have the same unit’s digit as the number itself. So their cyclicity is one. For example 156
^{5635}has unit’s digit 6. - Digits 4 & 9: Their cyclicity is two, 4
^{2n}and 9^{2n}have unit’s digit 4 and 9 respectively. 4^{2n+1}and 9^{2n+1}have unit’s digit 6 and 1 respectively. - Digits 2, 3, 7 & 8: Their cyclicity is four. 2
^{1}= 2, 2^{2}= 4, 2^{3}= 8, 2^{4}= 1**6**, 2^{5}= 3**2**. 3, 7 & 8 follows similar pattern.

Example: Find the Unit digit of 35^{87} + 93^{46}.

Since only unit’s digit is important, we need to find unit’s digit of 5^{87} + 3^{46}.

Unit’s digit of 5^{87} is 5 (as its cyclicity is 1).

3^{46} = 3^{4k+2}, so unit’s digit of 346 will be same as unit’s digit of 3^{2} = 9.

So the answer would be 5 + 9 = 14 => 4.

### 6. Remainder – Totient method

If *n* is prime number and *m* is co-prime to *n*, remainder of *m ^{n-1}* when divided by

*n*is always 1.

Example: Find remainder of 15^{26} when divided by 13.

15^{26} = 15^{24} × 15^{2}. Remainder of 15^{24} when divided by 13 would be 1.

So we need to find remainder of 15^{2} when divided by 13. This would be equal to 2^{2} when divided by 13, as 15 leaves remainder 2 when divided by 13.

So the answer would be 1 × 4 = 4.

If *n* isn’t prime, but co-prime to *m*, remainder of *m ^{D}* when divided by

*n*is always 1, where

*D*is totient of

*n*and given by-

*D = n(1 – 1/a)(1 – 1/b)(1 – 1/c) . . . , where a, b, c . . . are prime factors of n.*

D(26) = 26 × (1 – 1/2) × (1 – 1/13) = 12.

Example: Find remainder when 25^{25} is divided by 26.

Since totient of 26 is 12, m^{12} or m^{12k} when divided by 26 would give remainder 1.

25^{25} = 25^{24+1} would give remainder 25^{1} = 25.

You can find more such blogs here- Advance concepts and tricks of quantitative aptitude.

I would recommend you to practice solving questions from this section at Prepleaf – CAT Portal.