The digits of a positive integer n are four consecutive integers in decreasing order when . . .?

Question

The digits of a positive integer n are four consecutive integers in decreasing order when read from left to right (e.g. 9876, 8765, 7654, ..., 3210). What is the sum of the possible remainders when n is divided by 37?

Answer 1

Nice question!

the tricky thingie about 37 is that 37 * 3 = 111. That's a pretty number.

so suppose we were to get 9876.
111 * 90 = 9990
111 * 89 = 9990 - 111 = 9879, which is 9876 + 3.

likewise,
111 * 79 = 8769 = 8765 + 4
111 * 69 = 7659 = 7654 + 5
...
111 * 29 (last one) = 3219 = 3210 + 9

so the answer is 9+8+7+6+5+4+3 = 42

cheers!

Answer 2

Interesting question, but boring answer: 0. I'm assuming that the consecutive integers don't have to each be single digits, so that something like 21201918 works. You can check that 13121110 mod 37 is 22, and the remainders go up by 1 (including 28272625 mod 37 giving 0). By the time you hit 49484746 mod 37 you get 21, all possible remainders. Each element mod 37 has a distinct additive inverse (e.g., 22 + 15 is 0 mod 37), so that adding up all 37 possible values gives 0.

If you are only looking at 3210 up to 9876, the remainders are 28 up to 34, and the sum is 32 mod 37.

(Iluxa's observation about 111 is correct, but he should be summing negative numbers since, e.g., 9876 is 3 *less than* a multiple of 111. That gives -42 as the sum, which is congruent to -5 mod 37 and also 32 mod 37.)

Answer 3

It's 217. You divide 3210, 4321,... each by 37 and add up the remainders. Its 86 R28, 116 R29, 146 R30, 176 R31, 206 R32, 236 R33: 28+29+30+31+32+33=217

Answer 4

If a+j=p_1^a_1*p_2^a_2*...*p_r^a_r is the finest factorization of a+j, with each and each a_i>0, then a+j could have precisely (a_1+a million)*(a_2+a million)*...*(a_r+a million) divisors, jointly with a million and a+j. therefore contained in the present issue, we must have (a_1+a million)*(a_2+a million)*...*(a_r+a million) =6. this suggests r<=2. In case r=a million, we must have a_1=5 and subsequently a+j=p^5 for some best p. In case r=2, (a_1,a_2)=(a million, 2) or (2,a million), so hence we must have a+j=p*q^2 for some primes p and q. subsequently the sequence a+a million, a+2, .., a+N could comprise in reality integers which will properly be factored into between the varieties p^5 or p*q^2.

Answer 5

take a calculater and try