divisibility--help

[mitaliisrani]

1) How many 4 digit numbers can be formed using the digits 5,7,3,2 exactly once, such that the number formed is divisble by 11?

A) 16
B) 18
C) 24
D) 64
E) None of these

Ans:E



2) If a!-2a is a 3 digit integer divisible by 'a', what is the value of 2a+1?

A) 3 or 5
B) 6 or 7
C) 11 or 13
D) 3 or 13
E) 7 or 11


OA : C



3) How many 5 digit numbers formed using digits 0,1,2,3,4,5 exactly once, are divisible by 250?
A) 1
B) 2
C) 3
D) 4
E) 5


OA:B

[sanju09]

1) How many 4 digit numbers can be formed using the digits 5,7,3,2 exactly once, such that the number formed is divisble by 11?

A) 16
B) 18
C) 24
D) 64
E) None of these

Ans:E

No two sets of two of the given four digits are such that the difference in their sums could be 0, 11, 22 etc, hence no such number could possibly be formed using the digits 5,7,3,2 exactly once, such that the number formed is divisible by 11.

[spoiler]Choice (E) could better have been worded as 0.[/spoiler]

Post only one issue per thread, please

[sanju09]

2) If a!-2a is a 3 digit integer divisible by 'a', what is the value of 2a+1?

A) 3 or 5
B) 6 or 7
C) 11 or 13
D) 3 or 13
E) 7 or 11


OA : C

In factorial notations, the first 3-digit integer appears at 5! and disappears after 6!, and in both cases their form a! - a remains a 3-digit integer divisible by a, hence a can only take 5 or 6, and hence 2 a + 1 can only take [spoiler]11 or 13.

C[/spoiler]

[sanju09]

3) How many 5 digit numbers formed using digits 0,1,2,3,4,5 exactly once, are divisible by 250?
A) 1
B) 2
C) 3
D) 4
E) 5


OA:B

In the given constraints, there are a total of 5*5*4*3*2 ways of writing a 5-digit number, but that's not the question here. It asks that how many of those 5*5*4*3*2 ways are the 5-digit numbers divisible by 250. Remember, in the present case a number will be divisible by 250, if and only if its terminating three digits read 250. That way, we are left with arranging 2 out of the remaining three digits 1, 3, and 4 available to us, which could be done in [spoiler]3P2 = 6 ways.

Did I miss something here?[/spoiler]

[M811]

2) If a!-2a is a 3 digit integer divisible by 'a', what is the value of 2a+1?

A) 3 or 5
B) 6 or 7
C) 11 or 13
D) 3 or 13
E) 7 or 11


OA : C

In factorial notations, the first 3-digit integer appears at 5! and disappears after 6!, and in both cases their form a! - a remains a 3-digit integer divisible by a, hence a can only take 5 or 6, and hence 2 a + 1 can only take [spoiler]11 or 13.

C[/spoiler]

Can you please explain the solution in detail?

[sanju09]

2) If a!-2a is a 3 digit integer divisible by 'a', what is the value of 2a+1?

A) 3 or 5
B) 6 or 7
C) 11 or 13
D) 3 or 13
E) 7 or 11


OA : C

In factorial notations, the first 3-digit integer appears at 5! and disappears after 6!, and in both cases their form a! - a remains a 3-digit integer divisible by a, hence a can only take 5 or 6, and hence 2 a + 1 can only take [spoiler]11 or 13.

C[/spoiler]

Can you please explain the solution in detail?

Know that 0! = 1! = 1, 2! = 2, 3! = 6, 4! = 24, 5! 120, 6! = 720, 7! = 5040


And note that, the first 3-digit integer appears at 5! and disappears after 6!, and in both cases their form a! - a remains a 3-digit integer divisible by a, since 5! - 5 (= 120 - 5 = 115) is divisible by 5 and 6! - 6 (= 720 - 6 = 714) is divisible by 6, hence a can only take 5 or 6, and hence 2 a + 1 can only take [spoiler]11 or 13.

C[/spoiler]