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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.

 Nov 18, 2019
 #1
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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.

 Nov 18, 2019
edited by Guest  Nov 18, 2019
edited by Guest  Nov 18, 2019
edited by Guest  Nov 19, 2019
 #2
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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
edited by ElectricPavlov  Nov 18, 2019

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