Let three 3-digit integers abc, bca, cab when divided respectively by 10, 8 an

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Let three 3-digit integers abc, bca, cab when divided respectively by 10, 8 and 3 leave the same remainder 1. If $2 < b < 8$ , then a = ?

Options:

A . 3

B . 4

C . 5

D . Cannot be determined

Answer: Option A
:
A

abc leaves remainder of 1 when divided by $10 ⟹ c = 1$ .
bca leaves remainder of 1 on dividing by $8 ⟹ 4 b + 2 c + a = 8 m + 1$ and similarly a+b+c = 3n+1.
On solving we get 3b = 8m -1 - 3n $⟹$ m = 2, 5, 8,... or 3p-1 format.
Now, 4b+2+a = 8m+1 $⟹$ 4b+a = 8m-1
For p=3, m=8, 4b+a = 63 means b > 8. So, p cannot be greater than 2.
For p=2, m=5, 4b+a = 39, if a = 3, 7 means b $\geq$ 8 which is not possible.
So, p=1, m=2, 4b+a = 15, possible value of a = 3.
Hence, choice (a) is the right option.

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