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can P and R take?

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can P and R take?

by sanju09 » Fri Mar 05, 2010 6:07 am
A number P4571203R is divisible by 18. Which of the following values can P and R take, respectively?
(A) 1, 2
(B) 2, 3
(C) 3, 4
(D) 5, 9
(E) 6, 8
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by papgust » Fri Mar 05, 2010 7:42 am
Must be E

18 - 9*2.

For P4571203R to be divisible by 2, R must be even. Eliminate B, D.

For P4571203R to be divisible by 9, sum of digits must be divisible by 9. Apart from P and R, sum of digits is 22. So, adding P and R to the existing sum must be divisible by 9.

(A) 22 + 1 + 2 = 25 (Not divisible by 9). Eliminate
(C) 22 + 3 + 4 = 29 (Not divisible by 9). Eliminate
(E) 22 + 6 + 8 = 36 (Divisible by 9). RIGHT.
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by kstv » Fri Mar 05, 2010 7:57 am
IMO E
Similar to pagust. Eliminate B and D.
write the no on Scratch Pad
P4571203R strike out digit whose sum equal 9 eg 4&5, 7&2 you are left with 3&1 = 4
A 1 +2 = 3, 4+3 = 7 need 9 or its multiple
E 6,8 = 14 + 4 = 18 , voila!
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by fibbonnaci » Fri Mar 05, 2010 8:05 am
18 -> 9*2
For a number to be divisible by 9, the sum of the digits og the number must be divisble by 9.
In this case the sum is 22+P+R
So in order for the number to be divisible by 9, P+R when added to the number needs to make the number divisible by 9.
Only B and E left. Rest all are eliminated.
For the number to be divisble by 2, the last digit should be an even number.
So B does not satisfy the criteria.
Eliminated.
Left with E.
My answer E
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by sanju09 » Sat Mar 06, 2010 12:38 am
kstv wrote:IMO E
Similar to pagust. Eliminate B and D.
write the no on Scratch Pad
P4571203R strike out digit whose sum equal 9 eg 4&5, 7&2 you are left with 3&1 = 4
A 1 +2 = 3, 4+3 = 7 need 9 or its multiple
E 6,8 = 14 + 4 = 18 , voila!
I liked your voila!
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by sanju09 » Sat Mar 06, 2010 12:44 am
fibbonnaci wrote:18 -> 9*2
For a number to be divisible by 9, the sum of the digits og the number must be divisble by 9.
In this case the sum is 22+P+R
So in order for the number to be divisible by 9, P+R when added to the number needs to make the number divisible by 9.
Only B and E left. Rest all are eliminated.
For the number to be divisble by 2, the last digit should be an even number.
So B does not satisfy the criteria.
Eliminated.
Left with E.
My answer E


...and that's called an open secret!
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by fibbonnaci » Sat Mar 06, 2010 4:52 am
sanju09 wrote:
fibbonnaci wrote:18 -> 9*2
For a number to be divisible by 9, the sum of the digits og the number must be divisble by 9.
In this case the sum is 22+P+R
So in order for the number to be divisible by 9, P+R when added to the number needs to make the number divisible by 9.
Only B and E left. Rest all are eliminated.
For the number to be divisble by 2, the last digit should be an even number.
So B does not satisfy the criteria.
Eliminated.
Left with E.
My answer E


...and that's called an open secret!
I like the term- 'open secret' :)
Reminds me of the term- ' round circle' :P
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by sanju09 » Sat Mar 06, 2010 5:01 am
fibbonnaci wrote:
sanju09 wrote:
fibbonnaci wrote:18 -> 9*2
For a number to be divisible by 9, the sum of the digits og the number must be divisble by 9.
In this case the sum is 22+P+R
So in order for the number to be divisible by 9, P+R when added to the number needs to make the number divisible by 9.
Only B and E left. Rest all are eliminated.
For the number to be divisble by 2, the last digit should be an even number.
So B does not satisfy the criteria.
Eliminated.
Left with E.
My answer E


...and that's called an open secret!
I like the term- 'open secret' :)
Reminds me of the term- ' round circle' :P
[spoiler]...and round circle reminds me the remarks of a lean and thin man towards his terribly obese wife, which goes like...

My wife is a round circle and I am her radius. I'm all over inside her but she's nowhere inside me.[/spoiler]
The mind is everything. What you think you become. -Lord Buddha



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