Question

2 6 10 14 18 22 26 30 34 38 42 46 50 54 58 get any three no. and ans is 60 solve this question​

Answer 1

Answer:

Given numbers are of the form (2k-1)2 for k from 1 to 30.

If we take 3 numbers then they are (2a-1)2, (2b-1)2 and (2b-1)2 for some natural numbers a, b and c between 1 and 30 (both inclusive.

The numbers add up to

(2a-1)2+(2b-1)2+(2c-1)2

=(2a-1+2b-1+2c-1)2

=(2(a+b+c)-3)2

So, we get equation

(2(a+b+c)-3)2=60

=>2(a+b+c)-3=30

=>2(a+b+c)-3=30

=>2(a+b+c)=33

33 being odd, it cannot be double of sum of natural numbers

Ans.: No solution.

Another method:

Each of the numbers leaves remainder 2 when we divided by 4. If we add any two such numbers, we get multiple of 4. If we add another number leaving remainder 2 when divided by 4, new result would give remainder 2 when divided by 2. But 60 is multiple of 4 (remainder 0 when divided by 4).

One more method:

Let us divide the problem into half (I mean half of given numbers and total into half).

Given numbers become 1, 3, 5, …, 29 and total becomes 30.

If we can find 3 odd natural numbers adding to 30, doubling them would make the three numbers from original list adding up to 60.

So we need to find 3 odd numbers adding to 30. We know that odd+odd=even and even+odd=odd. So, three odd numbers cannot be added to 30.

Hence, no solution.

Step-by-step explanation: