Determine the number of five-digit integers (37abc) in base 10 such that each of the numbers (37abc), (37bca) and 37cab is divisible by 37.


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Question Number 19637 by Tinkutara last updated on 13/Aug/17

$Determine the number of five - digit$ $integers (37 a b c) in base 10 such that$ $each of the numbers (37 a b c), (37 b c a)$ $and 37 c a b is divisible by 37 .$
	Answered by Rasheed.Sindhi last updated on 14/Aug/17

	$37 abc = 37000 + abc$ $37 ∣ 37000 + abc$ $37 ∣ 3700 \Rightarrow 37 ∣ abc$ $\Rightarrow abc is multiple of 37$ $As a may be 0, abc may be$ $2 - digit or 3 - digit multiple$ $of 37 :$ $37 \times 0, 37 \times 1, 37 \times 2, . . ., 37 \times n$ $000, 037, 074, . . ., 37 n ⩽ 999 : 37 n$ $is last term GP .$ $n = [\frac{999}{37}] = 27$ $\overset{1}{000}, (\overset{n = 27}{037, 074, . . ., 37 n} ⩽ 999)$ $1 + 27 = 28$ $Total multiples are 28 having$ $property given in the question .$ $For example$ $37 = 037 \Rightarrow abc = 037 = 37 \times 1$ $bca = 370 = 37 \times 10, cab = 703 = 37 \times 19$ $So number 37037$ $Its variations : 37370 & 37703$ $37 ∣ 37037, 3 ∣ 37370, 37 ∣ 37703$
	Commented by Tinkutara last updated on 14/Aug/17

	$Thank you very much Sir!$
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