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if we open this ... in all (58-x)^7. there be 7 terms with powers of 58 and one term with only x all powers of 58 will be divisible by 58 we just need to check 45+44+43+42 = 174 also divisible by 58

if we open this ... in all (58-x)^7. there be 7 terms with powers of 58 and one term with only x all powers of 58 will be divisible by 58 we just need to check 45+44+43+42 = 174 also divisible by 58

if we open this ... in all (58-x)^7. there be 7 terms with powers of 58 and one term with only x all powers of 58 will be divisible by 58 we just need to check 45+44+43+42 = 174 also divisible by 58

remainder 0

I am not clear on this. The x terms all have power 7. How do you deal with that ? If 45+44+43+42 = 174 is divisible by 58, is 45^7 + 44^7 + 43^7...... also divisible by 58 ?

if we open this ... in all (58-x)^7. there be 7 terms with powers of 58 and one term with only x all powers of 58 will be divisible by 58 we just need to check 45+44+43+42 = 174 also divisible by 58

remainder 0

I am not clear on this. The x terms all have power 7. How do you deal with that ? If 45+44+43+42 = 174 is divisible by 58, is 45^7 + 44^7 + 43^7...... also divisible by 58 ?

If you expand any ploynomial such that (x - y ) ^n, you will always find x in all terms except last. So in this Q all you have to ensure is that sum of last terms is divisble by 58 or not. If yes then sum is divisible by 58

If you expand any ploynomial such that (x - y ) ^n, you will always find x in all terms except last. So in this Q all you have to ensure is that sum of last terms is divisble by 58 or not. If yes then sum is divisible by 58

I know that rule - binomial expansion. x will always be present in all terms except the last. But the last term will have power n. In context of current question, the last terms are not 45, 44, 43, and 42. But 45^7, 44^7, 43^7, and 42^7

I also know 45+44+43+42 = 174 is divisible by 58. But does that mean 45^7 + 44^7 + 43^7 + 42^7 is also divisible by 58 ?

If you expand any ploynomial such that (x - y ) ^n, you will always find x in all terms except last. So in this Q all you have to ensure is that sum of last terms is divisble by 58 or not. If yes then sum is divisible by 58

I know that rule - binomial expansion. x will always be present in all terms except the last. But the last term will have power n. In context of current question, the last terms are not 45, 44, 43, and 42. But 45^7, 44^7, 43^7, and 42^7

I also know 45+44+43+42 = 174 is divisible by 58. But does that mean 45^7 + 44^7 + 43^7 + 42^7 is also divisible by 58 ?

Hmmm .... you are right

I was trying another thing according to which

\(a^n + b^n + c^n + d^n\) is divisible by a + b + c + d if n is odd.

\(a^n + b^n + c^n + d^n\) is divisible by a + b + c + d if n is odd.

because \(a^n + b^n + c^n + d^n\) when n is odd will become (a + b + c + d) (a^n - a^n-1*b ....... )

I think there are additional constraints, i can think of atleast one.. that a,b,c,d need to be consecutive numbers. Coz it doesnt work with a=1, b=2, c=4. but works with 1,2,3 or 2,3,4 or 1,2,3,4.

But thanks buddy. Thats todays gain.. one new rule.

\(a^n + b^n + c^n + d^n\) is divisible by a + b + c + d if n is odd.

because \(a^n + b^n + c^n + d^n\) when n is odd will become (a + b + c + d) (a^n - a^n-1*b ....... )

I think there are additional constraints, i can think of atleast one.. that a,b,c,d need to be consecutive numbers. Coz it doesnt work with a=1, b=2, c=4. but works with 1,2,3 or 2,3,4 or 1,2,3,4.

But thanks buddy. Thats todays gain.. one new rule.

Yes you are right .... works only for consecutive numbers

So the expression is divisible by 2 and 29, and thus by 58.
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if you open it all terms with odd powers of 1.5 and 0.5 will cancel each other.....

for every Postive term with 7C0, 7C2, 7C4 and 7C6, we can take out 4*14.5 as a common factor .... (i did that on paper) and guess what is 4*14.5... its 58 ...

if you open it all terms with odd powers of 1.5 and 0.5 will cancel each other.....

for every Postive term with 7C0, 7C2, 7C4 and 7C6, we can take out 4*14.5 as a common factor .... (i did that on paper) and guess what is 4*14.5... its 58 ...

not that good... i think i just got lucky with this question .... there are faults in both of my approaches ... somehow got the right answer moral : try solving maths question while driving, your approach may be wrong, but you'll get the right answer