The digits of a positive integer n are four consecutive integers in decreasing order when read from left to right. What is the sum of the possible remainders when n is divided by 37?

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Solution

Dear student,

$L e t t h e d i g i t s o f n, r e a d f r o m l e f t t o r i g h t, b e a, a - 1, a - 2, a n d a - 3, r e s p e c t i v e l y, w h e r e a i s a n i n t e g e r b e t w e e n 3 a n d 9, i n c l u s i v e . b e c a u s e t h e m i m u m 4 d i g i t n u m b e r w i l l s t a r t f r o m 3000 i f d i g i t s a r e i n d e c r e a \sin g o r d e r . T h e n n = 1000 a + 100 (a - 1) + 10 (a - 2) + a - 3 = 1111 a - 123 = 37 (30 a - 4) + (a + 25), w e c a n s e e w h e n d i v i d i n g t h e n u m b e r b y 37, a + 25 i s t h e r e m a i n d e r w h i c h c a n b e a n y w h e r e b e t w e e n 0 t o 37 w h e r e 0 \leq a + 25 < 37 . T h u s t h e r e q u e s t e d s u m i s ({\sum (a + 25)}_{a = 3}^{9}) = (\sum_{a = 3}^{9} a) + 175 = 42 + 175 = 217 b e c a u s e a c a n r a n g e f r o m 3 t o 9 .$

answer is 217

Regards,

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