Questions & Answers

Question

Answers

a) 4

b) 7

c) 0

d) 8

Answer
Verified

By the prime-factorization method, we can find the factors of 54 and 87. So, we have:

Factors of 54 are: \[54=2\times 3\times 3\times 3=2\times 27\]

Factors of 87 are: $87=3\times 29$

So, the number 2575d568 to be divisible by 54 and 87, it should be divisible by 2, 27, and 29.

Since the unit’s digit of the number is an even number, therefore 2575d568 is always divisible by 2.

Since both 54 and 87 have 3 as their common factor. So, the eight-digit number 2575d568 must be at least divisible by 3.

To check whether a number is divisible by 3 we need to follow the divisibility rule by 3. It says that a number is divisible by 3 when the sum of digits is divisible by 3.

Example: $6561\Rightarrow 6+5+6+1=18$ is divisible by 3.

So, to find whether 2575d568 is divisible by 3 or not we need to check if the sum of digits of the number is divisible by 3 or not.

Therefore, we can write:

$\begin{align}

& \Rightarrow 2+5+7+5+d+5+6+8 \\

& \Rightarrow 38+d \\

\end{align}$

So, we need to put a one-digit value of d that makes $38+d$ divisible by 3.

i.e. d = 1, 4 or 7

If we add d=1, 4 or 7 in $38+d$ we get 39, 42 or 45 which are divisible by 3.

So, for d=1, 4 or 7, the number 2575d568 is divisible by 3. Hence the number would be divisible by 27.

But we need to check if for d=1, 4 or 7 the number 2575d568 is divisible by 29.

Put d=1, 4 or 7 in 2575d568, and check whether the number is divisible by 29. We get to know that for d=7, the number 2575d568 is divisible by 29.

Hence for d=7, the number 2575d568 is divisible by 54 and 87 both.

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