Question

Write one digit on each side of 92 to make a four-digit multiple of 45. How many different solutions does this problem have?

Answer 1

Answer:

45×65=2925

45×146=7920

there are only 2 multiples of 45 satisfying the condition

Answer 2

$Let \: the \: required \: number \: a92b$

$and \: it \: is \: multiple \: of \: 45$

$45 = 5 \times 9$

$5 \: and \: 9 \: are \: factors \:of \: a92b$

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$\underline{ Divisibility \:by \: 5 }}$

The given number is divisible by 5, only if the unit's digit is equal to '0' or '5' .

$\underline{ Divisibility \:by \: 9 }}$

The given number is divisible by 9 ,only the sum of the digits is multiple of 9 or divisible by 9 .

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$Unit \: place \: should \: be \: 0 \: are \: 5$

$\therefore b = 5 \: Or \: 0$

$and \: a = 2 \:or \: a = 7 \: respectively .$

Therefore.,

$\green { 2925 \: and \: 7920 \: are \: two }\\\green {four - digit \: multiple \: of \: 45 . }$

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