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555-605 (Medium)|   Algebra|      
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sherxon
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a, b, c and d are different digits. if a + b + c = 7 02 Jun 2021, 02:36
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76% (02:32) correct 24% (03:07) wrong based on 139 sessions

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a, b, c and d are different digits. If a + b + c = 7, \((a+b)^2\) = d and \(abc\neq{0}\), find \(\frac{c^2-c}{a+b}\)

A. 1
B. 2
C. 3
D. 4
E. 5
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D
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chetan2u
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a, b, c and d are different digits. if a + b + c = 7 02 Jun 2021, 03:54
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sherxon
a, b, c and d are different digits. If a + b + c = 7, \((a+b)^2\) = d and \(abc\neq{0}\), find \(\frac{c^2-c}{a+b}\)

A. 1
B. 2
C. 3
D. 4
E. 5
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So each of a, b, c and d are values from 0 to 9.
But \(abc\neq 0\), meaning none of them are 0.

Given that \((a+b)^2=d\): d is a square, so it can be 0, 1, 4 or 9.
But the minimum value of a and b is 1 and 2, so d has to be (1+2)^2=9.
Any increase in value of a and b will make d as 2-digit number.

Thus a and b are 1 and 2 in any order. => c=7-(1+2)=4.

Thus \(\frac{c^2-c}{a+b}\)= \(\frac{4^2-4}{1+2}\)=\(\frac{12}{3}\)=4


D
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a, b, c and d are different digits. if a + b + c = 7 02 Jun 2021, 04:06
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chetan2u
sherxon
a, b, c and d are different digits. If a + b + c = 7, \((a+b)^2\) = d and \(abc\neq{0}\), find \(\frac{c^2-c}{a+b}\)

A. 1
B. 2
C. 3
D. 4
E. 5


I am sure the question is missing something.

You can have hundreds of answers.
a+b+c=7...let a=-1, b=-2, so c=10 and d=(-1-2)^2=9
c^2-c/(a+b)=90/(-3)=-30
a+b+c=7...let a=1, b=2, so c=4 and d=(1+2)^2=9
c^2-c/(a+b)=12/(3)=4.

There is almost no role of d except telling that (a+b)^2 cannot be c.
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I think everything is clear ;)

Definition of digit from Merriam-Webster:
1a: any of the Arabic numerals 1 to 9 and usually the symbol 0
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a, b, c and d are different digits. if a + b + c = 7 02 Jun 2021, 04:11
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sherxon
chetan2u
sherxon
a, b, c and d are different digits. If a + b + c = 7, \((a+b)^2\) = d and \(abc\neq{0}\), find \(\frac{c^2-c}{a+b}\)

A. 1
B. 2
C. 3
D. 4
E. 5


I am sure the question is missing something.

You can have hundreds of answers.
a+b+c=7...let a=-1, b=-2, so c=10 and d=(-1-2)^2=9
c^2-c/(a+b)=90/(-3)=-30
a+b+c=7...let a=1, b=2, so c=4 and d=(1+2)^2=9
c^2-c/(a+b)=12/(3)=4.

There is almost no role of d except telling that (a+b)^2 cannot be c.

I think everything is clear ;)

Definition of digit from Merriam-Webster:
1a: any of the Arabic numerals 1 to 9 and usually the symbol 0
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Hi

You are correct. I for some reasons read digit as numbers.

Let me edit my answer.
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maheshhchaudhari
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a, b, c and d are different digits. if a + b + c = 7 02 Jun 2021, 04:33
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Is answer 4 that is D ?
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a, b, c and d are different digits. if a + b + c = 7 02 Jun 2021, 04:38
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maheshhchaudhari
Is answer 4 that is D ?
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Your answer is correct, but it would be useful for others if you could post how you arrived at your answer ))
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a, b, c and d are different digits. if a + b + c = 7 02 Jun 2021, 04:42
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Within two minutes, I rush to assume a = 1, b = 2, c = 4 as d is just to confirm whether a and b are correct.
Substitute values in C^2-C/a+b give 4.
Later on I verified that (1+2)^2 = 9 = d.
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a, b, c and d are different digits. if a + b + c = 7 04 Jun 2021, 23:29
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chetan2u
sherxon
a, b, c and d are different digits. If a + b + c = 7, \((a+b)^2\) = d and \(abc\neq{0}\), find \(\frac{c^2-c}{a+b}\)

A. 1
B. 2
C. 3
D. 4
E. 5

So each of a, b, c and d are values from 0 to 9.
But \(abc\neq 0\), meaning none of them are 0.

Given that \((a+b)^2=d\): d is a square, so it can be 0, 1, 4 or 9.
But the minimum value of a and b is 1 and 2, so d has to be (1+2)^2=9.
Any increase in value of a and b will make d as 2-digit number.

Thus a and b are 1 and 2 in any order. => c=7-(1+2)=4.

Thus \(\frac{c^2-c}{a+b}\)= \(\frac{4^2-4}{1+2}\)=\(\frac{12}{3}\)=4


D
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Wonderful explanation. thank you very much
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a, b, c and d are different digits. if a + b + c = 7 04 Jun 2021, 23:53
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[quote="maheshhchaudhari"] keeping a=4 b=2 c=1 answer coming 0

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a, b, c and d are different digits. if a + b + c = 7 29 Nov 2024, 23:21
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