Question

Find the cleast number which
increased
by 3, is divisible by 36,40 and
64​

Answer 1

Answer:

2877

Step-by-step explanation:

Hence, 2877 is the least number which when increased by 3 is divisible by 36, 40 and 64.

Answer 2

Solution :-

To find the least number which when increased by 3 is divisible by 36, 40 and 64, we first have to find the L.C.M. of 36, 40 and 64

Prime Factorization of numbers  

36 = 2*2*3*3

40 = 2*2*2*5

64 = 2*2*2*2*2*2

L.C.M. of 36, 40 and 64 = 2*2*2*2*2*2*3*3*5

= 2880

So, 2880 is the number which is already increased by 3.

Therefore, the number we need is 3 less than 2880

So, the required number is 2880 - 3 = 2877

Hence, 2877 is the least number which when increased by 3 is divisible by 36, 40 and 64.