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Find the values of
x and
y such that the number
56129137X51Y is divisible by 88
(a) 2,2
(b) 4, 2
(c) 4,4
(d) none of these

$1$ . Find the values of $x$ and $y$ such that the number $56129137 \mathrm{X} 51 \mathrm{Y}$ is divisible by $88$ $\newline$ (a) $2$ , $2$ $\newline$ (b) $4$ , $2$ $\newline$ (c) $4$ , $4$ $\newline$ (d) none of these

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Write equations of parabolas in vertex form using properties

Full solution

Q.

1

. Find the values of

x

and

y

such that the number

56129137 \mathrm{X} 51 \mathrm{Y}

is divisible by

88

\newline

(a)

2

,

2

\newline

(b)

4

,

2

\newline

(c)

4

,

4

\newline

(d) none of these

Divisibility by $88$ : To be divisible by $88$ , the number must be divisible by both $8$ and $11$ , since $88 = 8 \times 11$ .
Check Divisibility by $8$ : First, check divisibility by $8$ . The last three digits must be divisible by $8$ . So, $51Y$ must be divisible by $8$ .
ext{Y} = $2$ for Divisibility by $8$ : Try ext{Y} = $2$ , then $512$ is divisible by $8$ because $512 ext{ ÷ } 8 = 64$ , which is a whole number.
Check Divisibility by $11$ : Now, check divisibility by $11$ . The difference between the sum of the digits in the odd positions and the sum of the digits in the even positions must be either $0$ or a multiple of $11$ .
Add Digits in Odd Positions: Let's add the digits in the odd positions: $5 + 1 + 9 + 3 + X + 5 + 2 = 25 + X$ .
Add Digits in Even Positions: Now, add the digits in the even positions: $6 + 2 + 1 + 7 + 1 + Y = 17 + Y$ .
Difference for Divisibility by $11$ : The difference is $(25 + X) - (17 + Y)$ . For divisibility by $11$ , this difference must be $0$ or a multiple of $11$ .
Try $X = 4$ and $Y = 2$ : If we try $X = 4$ and $Y = 2$ , the difference is $(25 + 4) - (17 + 2) = 29 - 19 = 10$ , which is not a multiple of $11$ .

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