Maths: Practice Paper on Number System for CLAT

Odd places digits sum= 1+x+9+7 =17+x

Even places digits sum = 5+6+3 =14

Difference between the two sums = 17+x – 14 = 3+x

As per divisibility rule (3+x) should be divisible by 11, therefore 11= x+3

x = 11 – 3 = 8

As we divide 2016 by 19 we get the remainder 2, that means when we subtract 2 from 2016 the number gets completely divisible.

*For solving unit digit questions we don’t have to use whole numbers of products, only significant digits are of units place.

For such questions multiply all the digits at units place and what you get in the units place of the new number is the digit at units place of multiple of numbers given in question.*

Multiple of digits at units place 3*6*4*9 = 18*36

Now here you might find yourself stuck with big numbers but it’s not that complicated. For such big numbers, you can again apply the same trick and it won’t change results i.e., 8*6 = 48

Thus the unit digit of (463 * 756 * 414 * 679) is 8

Given number: 47* 75 * 1111

Here we see other than 4, 7 and 11 are prime numbers but we also know that 4 is 2*2 which is a prime number. Thus we can rewrite the numbers as

(22 )7  * (7)5  * (11)11  = 214 *75 *1111

The no. of prime numbers would be the sum of numbers in power because that many prime numbers are there in the number ie., 14+5+11 = 30.

thus we see here 77*110 gives 8470 and 18 becomes remainder. Thus the nearest natural number is 8470.

This is again a remainder question with just twisted language. If you find such questions in exam read th question twice-thrice and you would get the trick.

If 10000 is the smallest number and adding 31-18= 13 would make it divisible by 31.

If 99999 is the largest number and subtracting 81would make it divisible by 91.

Suppose a is the number being divided and b be the quotient

a= 56 *b + 29

= 8*7*b + 8*3 +5

= 8 ( 7*b + 3) +5

We need to divide a by 8 for fiding out the remainder, in the above explanation first part 8( 7*b + 3) have a factor 8 but 5 would still remain as remainder.

The first 3 digit number that is divisible by 6 is 102 and last would be 996 (you can find these numbers using smallest and largest methods we discussed earlier ).

This becomes an A.P. with a= 102, d=6 and l=996

l= a + (n-1)d

996 = 102 + (n-1) 6

After solving above equation we get n=150. Therefore 150 3 digit numbers which are divisible by 6.