**#Problem 12.1**

if a^2+b^2 is a multiple of 7^2009 then prove that ab is a multiple of 7^2010.[USA Math Talent Search]

**Solution scheme and approach**

a^2+b^2=7^2009*k

As right hand side is divisible by 7 left hand should be divisible by 7.

Now we can deduce the the following table

**Number-----Square-------remainder (modulo 7)**

0-------------0-------------0

1-------------1-------------1

2-------------4-------------4

3-------------9-------------2

4------------16-------------2

5------------25-------------4

6------------36-------------1

From the table it's clear that as sum of two numbers (>=1) gives 7 so both of them should be divisible by 7.

Let a=7a1,b=7b1.

=>a^2+b^2=7^2009*k

a1^2+b1^2=7^2007*k

Likewise, let a1=7a2 and b1=7b2

a2^2+b2^2=7^2005*k

=>a=7^2*a1 and b=7^2*b1

and so on ...................

at last we get

a1004^2 + b1004^2=7k

=>a=7^1005*a1 and b=7^1005*b1

=>ab=7^2010*(a1b1)

**=>ab is a multiple of 7^2010. Q.E.D.**

**#Problem 12.2**

A set consists of 143 natural numbers each of them is a perfect cube. If they are divided by 13, how many maximum numbers within the set are possible, such that one will get the same reminder in each case.

(a)11, (b)31, (c)33, (d)36, (e)29 [AIMCAT]

When any cube(0^3,1^3,2^3,...,12^3) divides by 13 it generates 5 different numbers 0,1,5,8,12. Now at least [143/5]+1=29 numbers are possible which will generate same remainder.

Hence, option (e) is the answer

**Solution scheme and approach**When any cube(0^3,1^3,2^3,...,12^3) divides by 13 it generates 5 different numbers 0,1,5,8,12. Now at least [143/5]+1=29 numbers are possible which will generate same remainder.

Hence, option (e) is the answer

why are you making things look difficult?

ReplyDeletemost of the things on your blog are completely irrelevant as far as CAT is concerned.