Divisible by this year? v2015 (Part 1: Happy New Year!)

Number Theory Level 4

Let $a_{1}$ , $a_{2}$ , $a_{3}$ , $\dots$ , $a_{2013}$ , and $a_{2014}$ be $2014$ positive distinct prime integers where $a_i < a_{i + 1}$ for all positive integer $i < 2014$

Find the least possible value of $a_1$ if

${ { a }_{1 } }^{ 2}+ { { a }_{2 } }^{ 2}+{ { a }_{3 } }^{ 2} + \dots + { { a }_{2013 } }^{ 2}+ { { a }_{2014 } }^{ 2}$

is divisible by $2015$

Please show your solutions! Thanks!!!

1 solution

U Z
Jan 3, 2015

Except 2 all the other prime numbers are odd

If we take all the numbers odd prime then the result cannot be divisible by 2015 as the sum would be even

thus for the sum to be divisible by an odd number any one number should be even

Its not necessary that the number should be odd to be divisible by 2015 as 4030 is also divisible by 2015...

Vighnesh Raut - 6 years, 5 months ago

4030 is an even number

U Z - 6 years, 5 months ago

oh i mean that it isn't necessary for a number to be divisible by 2015 if it is odd , it can be even too....

Vighnesh Raut - 6 years, 5 months ago

@Vighnesh Raut – digits at the end of prime numbers - 2 , 3 , 5 , 7

now their square(digits at unit place) - 4 , 9 , 5 , 9

expect 5 none is a prime number and for the sum divided by 2015 primilarily we should have 0 as its unit's place( others are secondary)

now take the sums of the squares of these prime number ( leave here 2) no sum have 0 as a result in its units place

U Z - 6 years, 5 months ago

@U Z – oh i c......thnx ..

Vighnesh Raut - 6 years, 5 months ago

Divisible by this year? v2015 (Part 1: Happy New Year!)

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