Question

(75-a)(75-b)(75-c)(75-d)(75-e)=2299 find the value of a+b+c+d if (a,b,c,d,and e) are distinct integers a.330 b.300 c.390 d.530

Answer 1

$<b > Hello It's Jeet Here For Your Help$

$Here \: is \: your \: answer$

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If we factorize 2299 then we get.....

2299= 11 X 11 X 19 X 1 X 1

We can also write the above expression as 

2299= 11 X (-11) X 19 X (-1) X 1 

So we can write (75-a)(75-b)(75-c)(75-d)(75-e) = 2299 as

(75-64)(75-86)(75-56)(75-76)(75-74) = 2299

[ As 75-11 = 64 , 75-(-11) = 86 , 75-19 = 56 , 75-(-1) = 76 , 75-1 = 74 ]

So , a + b + c + d + e = 64 + 86 + 56 + 76 + 74 = 356

But here they wanted only the sum of 4 terms.

So, subtracting one of the terms respectively from the total we can get 292, 270, 300, 282, 280

So correct option is 300.

Hope it is clear to you

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$Hope \: the \: above \: answer \: helps$

Answer 2

2299 = 11×11×19×1×1=11× −11×19× −1×1=

Two of the terms in the given expression should equal to

1. As all the digits are distinct, two of the terms should be negative.

One possible solution = (75 - 64)(75 - 56)(75 - 86)(75 - 74)(75 - 76)

Then a + b + c + d + e = 64 + 56 + 86 + 74 + 76 = 356

But as the sum of only 4 terms was asked, we have to subtract one term.

So given answer can be one of 292, 306, 270, 282, 280