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Correct Relation

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Correct Relation

by GMATSUCKER » Wed Mar 03, 2010 11:55 am
Q: Which one of the following is not the correct relation ;

a)25^3 + 38^3 + 87^3 =90^3

b)3^3 + 4^3 + 5^3 = 6^3

c)17^3 + 29^3 = 57^3

d)1^3 + 6^3 + 8^3 = 9^3

e)All of these

OAC

Now, is there a short cut formula to find the relation ?
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by ajith » Wed Mar 03, 2010 12:22 pm
GMATSUCKER wrote:Q: Which one of the following is not the correct relation ;

a)25^3 + 38^3 + 87^3 =90^3

b)3^3 + 4^3 + 5^3 = 6^3

c)17^3 + 29^3 = 57^3

d)1^3 + 6^3 + 8^3 = 9^3

e)All of these

OAC

Now, is there a short cut formula to find the relation ?
I cannot think about a formula but, can certainly think about a trick

a)25^3 + 38^3 + 87^3 =90^3

Last digit of 25^3 is 5; 38^3 is 2 and 87^3 is 3 => the sum ends in zero so does 90^3; so cannot rule out this option

b)3^3 + 4^3 + 5^3 = 6^3

This we can calculate 125+64 +27 = 216

c)17^3 + 29^3 = 57^3

Last Digit of 17^3 is 3; last digit of 29^3 is 9; sum ends in 2 but 57^3 must end in 3 => this equation cannot be true

d)1^3 + 6^3 + 8^3 = 9^3

We can calculate 512+216+1 = 729

Hence C
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by GMATSUCKER » Wed Mar 03, 2010 12:39 pm
ajith wrote:
GMATSUCKER wrote:Q: Which one of the following is not the correct relation ;

a)25^3 + 38^3 + 87^3 =90^3

b)3^3 + 4^3 + 5^3 = 6^3

c)17^3 + 29^3 = 57^3

d)1^3 + 6^3 + 8^3 = 9^3

e)All of these

OAC

Now, is there a short cut formula to find the relation ?
I cannot think about a formula but, can certainly think about a trick

a)25^3 + 38^3 + 87^3 =90^3

Last digit of 25^3 is 5; 38^3 is 2 and 87^3 is 3 => the sum ends in zero so does 90^3; so cannot rule out this option

b)3^3 + 4^3 + 5^3 = 6^3

This we can calculate 125+64 +27 = 216

c)17^3 + 29^3 = 57^3

Last Digit of 17^3 is 3; last digit of 29^3 is 9; sum ends in 2 but 57^3 must end in 3 => this equation cannot be true

d)1^3 + 6^3 + 8^3 = 9^3

We can calculate 512+216+1 = 729

Hence C
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by kstv » Thu Mar 04, 2010 7:48 am
a)25^3 + 38^3 + 87^3 =90^3

b)3^3 + 4^3 + 5^3 = 6^3

c)17^3 + 29^3 = 57^3

d)1^3 + 6^3 + 8^3 = 9^3


a³+b³+c³ < (a+b+c)³
This rule is not violated except in C where 17+29 < 57
57 = 17 + 29 + 11 so 57³ has to be > just 17³+29³

If the above is confusing consider a²+b² = c²
always a+b > c
eg. 3²+4²=5², 8² has to be > 3²+4²
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by Ian Stewart » Thu Mar 04, 2010 10:23 am
GMATSUCKER wrote:
c)17^3 + 29^3 = 57^3
You do not need any inequality or units digit relationships to see that this cannot be true: on the left side of the equation, we're adding two odd numbers. That must give us an even number, and 57^3 is not even.
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by melvinsgmat » Thu Mar 04, 2010 11:29 am
Ian Stewart wrote:
GMATSUCKER wrote:
c)17^3 + 29^3 = 57^3
You do not need any inequality or units digit relationships to see that this cannot be true: on the left side of the equation, we're adding two odd numbers. That must give us an even number, and 57^3 is not even.
Hi Ian,

Can you explain that point in detail. Do you mean that I need to add the the numbers on the left side to get a even number and thus it satisfies the relation.


taking the case of a

25+38+87=150 = LHS and it is a even number . RHS=90 that is also even and thus relation fine.

3+4+5=12= LHS and it is even also and RHS=6 is also even

17+29=46=LHS and even and RHS=57 and it is not even

1+6+8=15=LHS and odd and RHS=9 and is odd

Please clarify. thanks in advance.
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by Ian Stewart » Thu Mar 04, 2010 6:42 pm
melvinsgmat wrote:
Ian Stewart wrote:
GMATSUCKER wrote:
c)17^3 + 29^3 = 57^3
You do not need any inequality or units digit relationships to see that this cannot be true: on the left side of the equation, we're adding two odd numbers. That must give us an even number, and 57^3 is not even.
Hi Ian,

Can you explain that point in detail. Do you mean that I need to add the the numbers on the left side to get a even number and thus it satisfies the relation.


taking the case of a

25+38+87=150 = LHS and it is a even number . RHS=90 that is also even and thus relation fine.

3+4+5=12= LHS and it is even also and RHS=6 is also even

17+29=46=LHS and even and RHS=57 and it is not even

1+6+8=15=LHS and odd and RHS=9 and is odd

Please clarify. thanks in advance.
I'm not sure I understand your question, but I will try to explain: I was only pointing out that 17^3 is odd, and that 29^3 is odd. So when we add them, we must get an even number; 17^3 + 29^3 could not possibly be equal to 57^3, which is an odd number. The question asks us which relation is not correct, and this gives us a quick way to see that the relation in answer choice C is not correct. I wasn't doing any calculations using numbers at all; I was just considering whether the sum ought to be even or odd.
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by GMATSUCKER » Fri Mar 05, 2010 4:06 am
Thanks to everybody for providing their insights. Ajith's way was quite good : checking the unit's digit which is a surefire technique in case of doubt ; KSTV's way was also good using the formula method , but Ian's method was the fastest :- checking odd/even. With these three methods I don't think any student can ever make an error.
I thank all for their participation & look forward to their future participation.
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by STEVEN SPIELBERG » Sat Mar 13, 2010 11:24 am
C . Great ways to solve it !
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by Aman verma » Sat Mar 13, 2010 11:57 am
Yes, good approach suggested by everybody . I would also like to know whether there is any formula based approach to solve these type of problems !!
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by birdistheword » Mon Apr 19, 2010 2:44 am
Ian's suggested approach is best and fastest I suppose.

I used the approach to equate last digits. Though I would like to add a point that might help in saving a few precious seconds

In such questions I always tend to start from choice E. Since in this case you need to evaluate only 4 choices, if you had started from D, you would have found the answer in C.
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