The sum of squares of 3 positive integer is 323. if the sum of the squares of the two numbers is twice the third, their product is
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Heya User,
-->Ur question, ummm.. I find it somewhat wrong..
--> let's do it my way ..
--> Let the no.s be our favorite :-> "a","b","c"
--> ATQ --> a² + b² + c² = 323
We know, Square of any no. ends with --> 0,1,4,5,6,9
Now, if uh see it that way :-> when the square of any no. is divided by
10, the remainder is 0,1,4,5,6,9
--> However, 323 leaves remainder 3 when divided by 10.
--> Hence, all the three squares end with a "1" at the last
--> This implies, the no.s ( a, b, c ) ends with {1,9}...
Now, also 19² = 361 > 323
Hence, our only options are :-> 1 , 9, 11 for a, b, c
Again, a few calculations would help you to find that :->
-------> a = 9 || b = 11 || c = 11 ;
And we're done... Their product is 121*9 = 1089 ...xD
Hope you'll try looking at the questions a new way now..
-->Ur question, ummm.. I find it somewhat wrong..
--> let's do it my way ..
--> Let the no.s be our favorite :-> "a","b","c"
--> ATQ --> a² + b² + c² = 323
We know, Square of any no. ends with --> 0,1,4,5,6,9
Now, if uh see it that way :-> when the square of any no. is divided by
10, the remainder is 0,1,4,5,6,9
--> However, 323 leaves remainder 3 when divided by 10.
--> Hence, all the three squares end with a "1" at the last
--> This implies, the no.s ( a, b, c ) ends with {1,9}...
Now, also 19² = 361 > 323
Hence, our only options are :-> 1 , 9, 11 for a, b, c
Again, a few calculations would help you to find that :->
-------> a = 9 || b = 11 || c = 11 ;
And we're done... Their product is 121*9 = 1089 ...xD
Hope you'll try looking at the questions a new way now..
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