can P and R take?

[sanju09]

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

[papgust]

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.

[kstv]

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!

[fibbonnaci]

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

[sanju09]

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!

[sanju09]

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!

[fibbonnaci]

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'

[sanju09]

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'

[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]