Question

find the largest number that divides 2053 and 967 and leaves a remainder of 5 and 7

Answer 1

it is given that on dividing 2053 by the required number there is a remainder of 5 .this mean that 2053 - 5 = 2048 is exactly divisible by required number . similarly,
967 - 7 = 960
the required number is the largest number satisfying the above property .
therefore ,it is the HCF of 2048 & 960
HCF of 2048 & 960 is 64 
hence required number is 64

Answer 2

So, the number should leave a remainder of 5 when it divides 2053. That means, that number, say 'x' will perfectly divide 2053-5 = 2048.

The same number 'x' will leave a remainder of 7 when it divides 967. So, it perfectly divides 967-7 = 960.

Since you need the largest number, now, we find out the HCF (Highest Common Factor) of 2048 and 960.

Factorize them.

2048 =
${2}^{11}$
960 =
${2}^{5} \times 3 \times 10$
Common value is 2 raised to power five, that is, 32.

Thus, the answer is 32.