Math, asked by alex2565, 9 months ago

let a,b,c be distinct integers such that b+c-a, c+a-b and a+b-c are all perfect squares. what is the largest possible value of a+b+c smaller than 100​

Answers

Answered by priyanshu2776
0

Step-by-step explanation:

st.1 as a,b,c>0 thus all three of them must be less than 5. because if any of the number a,b,c becomes 5. then the value of a+b+c will be greater than 7.

thus three numbers will be less than 5. so possible values available for a,b,c are 1,2,3,4. of these possible values only 1,2,4 results in the some of 7. thus product of a,b,c will be 1*2*4=8

st.2 again, if any of the number a,b,c becomes 5 or bigger then the sum of ab+bc+ca will be greater than 5. hence a,b,c will be less than 5.

from the possible set of values of a,b,c only 1,2,4 results in the sum of 14. hence product of a,b,c will be 1*2*4

For S1, let's try a few possibilities. 1 + 2 + 3 = 6 is the smallest, and that doesn't work. 1 + 2 + 4 does, so that's a possibility. 1 + 2 + 5 is too big, so that (or anything 1 + 2 + a number bigger than 5, for that matter) is out. The smallest possibility without 1 involved is 2 + 3 + 4 = 9, which is also too big ... so we must have 1, 2, and 4. Hence abc = 1 * 2 * 4 = 8, so S1 is SUFFICIENT.

For S2, algebra doesn't seem to help, so let's use number properties instead.

ab + bc + ca = 14, so ab + bc + ca = even. We can only have an EVEN sum if AT LEAST one of ab, bc, and ca is even. (If they're all odd, we'd have odd + odd + odd, which is NOT even.) If at least one of ab, bc, and ca is even, then at least one of a, b, and c is even. Now let's suppose that ONLY one of them is even: in other words, that a is even, b is odd, and c is odd. Then ab + bc + ca = evenodd + oddodd + odd*even = even + odd + even = odd. So this isn't possible either!

Hence, we know that at least TWO of a, b, and c are even. Now let's suppose that ALL THREE are even. Using the SMALLEST possible even a, b, and c, we'd have 2 * 4 + 2 * 6 + 4 * 6, which is too big ... so they can't all be even!

So we know we have two even and one odd.

1 * 2 + 1 * 4 + 2 * 4 = 14 works, so we could have a * b * c = 1 * 2 * 4

2 * 3 + 2 * 4 + 3 * 4 = 26, which is too big ... so anything other than 1, 2, and 4 is too big. Hence a * b * c = 8, and S2 is ALSO sufficient.

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