Let a, b, c, d be four consecutive positive integers, in that order, such that a is divisible by 5, b is divisible by 7, c is divisible by 9, and d is divisible by 11. Find the minimum value of a + b + c + d.

Guest Nov 18, 2019

#1**+2 **

There maybe many solutions to your probelem. You didn't state the number of digits in each integer.

For example, here are 4 consecutive 5-digit integers that satisfy all the congruences:

12130 mod 5=0, 12131 mod 7=0, 12132 mod 9=0, 12133 mod 11=0

12130 + 12131 + 12132 + 12133 = 48,526

**P.S. Here are 4 consecutive 4-digit integers that also satisfy all the congruences. According to computer search it seems to indicate these are the smallest 4 consecutive integers:**

**1735 mod 5 =0, 1736 mod 7=0, 1737 mod 9 =0, 1738 mod 11 =0**

**1735 + 1736 + 1737 + 1738 = 6,946.**

**(1735, 5200, 8665, 12130, 15595, 19060, 22525, 25990, 29455, 32920, 36385, 39850, 43315, 46780, 50245, 53710, 57175, 60640, 64105, 67570, 71035, 74500, 77965, 81430, 84895, 88360, 91825, 95290, 98755)>>Total = 29 numbers up to 100,000 that satisfy the 4 congruences. However, there are 289 numbers up to 1,000,000 that satisfy the 4 congruences. Each number begins with mod 5=0. Then 1 is added to the follwing 3 numbers. For example:5200 mod 5, 5201 mod 7, 5202 mod 9 and 5203 mod 11 =0.**

Guest Nov 18, 2019

edited by
Guest
Nov 18, 2019

edited by Guest Nov 18, 2019

edited by Guest Nov 19, 2019

edited by Guest Nov 18, 2019

edited by Guest Nov 19, 2019

#2**0 **

Nice work!

....I have no idea if your answer is correct...There may be multiple answers for a b c d ...BUT question asks for MINUMUM value of a+b+c+d

ElectricPavlov
Nov 18, 2019