A positive integer is divisible by 3 if and only if the sum

n will be divisible by 3 if (1+k+2+k+2+4=2k+9) is divisible by 3.
Since 9 is divisible by 3, we have to find the values of k (k is single digit) for which 2k is divisible by 3.
k can be 3,6 or 9, hence possible values of n can be 132324 or 162624 or 192924.
Hence number of possible values of n is 3. Answer is 2

I think it is actually 4. You did not include 0 as a value of k.

Edit: 4 as in 4 values of k, which is really answer choice 3.

ANSWERS

1. C
2. D
3. C

SOLUTIONS

1.

For n to be divisible by 10, y has to be zero. So, n = 4x130

As suggested by Karishma above, when doing the divisibility test for 3, we don't need to add all digits: 4+x+1+3+0. We can simply add the digits that are non-multiples of 3 or whose sum is not a multiple of 3. So, in this case, we need to only add 4+x+1 = 5+x

For what values of x will 5+x be divisible by 3? For x=1, 4, 7.

So, in all there are only 3 possible numbers that will be divisible by both 3 and 10:
41130, 44130, 47130

(C) is the answer.

2.
For n to be divisible by 5, y could either be 0 or 5.

Case 1: y= 0

So, n= 4x130

Using the divisibility rule for 3, we find that possible values of x are 1, 4, 7. However, it is given that x>1. So, in total, only 2 values of x are possible.

Case 2: y=5

So, n= 4x135

Here sum of 1+3+5 = 9, a multiple of 3. So, we need to only consider the digits 4 and x for the divisibility test.

Possible values of x are 2, 5, 8. So, 3 values are possible.

Thus, total possible values of n are 5:

44130, 47130
42135, 45135, 48135


3.
For n to be divisible by 2, y= 0, 2,4,6,8. However, it is given that y>2. So, allowed values of y: 4, 6, 8

Case 1: y= 4

So, n= 4x134
Possible values of x = 0,3,6,9. Therefore, 4 values.

Case 2: y=6

So, n= 4x136
We need to only consider the digits 4, x, 1 for the divisibility rule (since 3 and 6 are already multiples of 3).
Possible values of x= 1,4,7. Therefore, 3 values

Case 3: y=8

So, n= 4x138
We need to consider only the digits 4 and x for the divisibility rule (since 3 and 8+1 are already multiples of 3).
Possible values of x = 2, 5, 8. Therefore, 3 values

Total values of n possible = 4+3+3 = 10

Please find below the complete list of divisibility rules:

Divisible by:
2 IF: The last digit is even (0,2,4,6,8)
3 IF: The sum of the digits is divisible by 3
4 IF: The last 2 digits are divisible by 4
5 IF: The last digit is 0 or 5
6 IF: The number is divisible by both 2 and 3
7 IF: If you double the last digit and subtract it from the rest of the number and the answer is: 0, or divisible by 7
(Note: you can apply this rule to that answer again if you want) Eg: 672 (Double 2 is 4, 67-4=63, and 63÷7=9) Yes
8 IF: The last three digits are divisible by 8
9 IF: The sum of the digits is divisible by 9
10 IF: The number ends in 0
11 IF: If you sum every second digit and then subtract all other digits and the answer is: 0, or divisible by 11
Eg: 3729 ((7+9) - (3+2) = 11) Yes; 25176 ((5+7) - (2+1+6) = 3) No
12 IF: The number is divisible by both 3 and 4

Cheers,
Rajat

To further clarify on this: It is not a one time operation but one needs to keep subtracting double of last digit till it results in a 2 digit number.
Example: 1603 -> 160-2(3)=154 -> 15-2(4)=7, so 1603 is divisible by 7.

Look up here for more divisibility checks: https://www.savory.de/maths1.htm

1k2k24

Taking sum of the numericals = 1 + 2 + 2 + 4 = 9

We require the values of K such that they are also divisible by 3

102024

132324

162624

192924

Answer = 4 = C

LCM of 3 & 10 = 30

To make 4x13y divisible by 30, y has to be 0

Now try to find values for x

4 + 1 + 3 + 0 = 8

We require the addition of digits to be divisible by 3,

so x can be 1 or 4 or 7

No. created would be:

41130

44130

47130

Answer = 3 = C

To make no. divisible by 3, sum of all digits should be divisible by 3

To make no. divisible by 5, y has to be either 0 OR 5

When y = 0, x can be 4 or 7 (1 has to be ignored as x>1)

44130

47130

When y = 5, x can be 2, 5 or 8

42135

45135

48135

Total 5 combinations possible

Answer = 5 = D

When y>2, it may be 4 , 6 or 8

When y = 4, x may be 0, 3, 6, 9

When y = 6, x may be 1, 4, 7

When y = 8, x may be 2, 5, 8

Total combinations = 4 + 3+ 3 = 10

Answer = C

Thanks Bunuel! I think I did. But I think I didnt make full use of the subject line. Thanks for sharing the positing rules also. Will follow them exactly now.

Hi Karishma,

So question for you here -- if the above example ended at 3627598172 -- we would be holding 5. Does that mean that the number is NOT divisible by 3?

Yes, then the number would not be divisible by 3 and on dividing by 3 the remainder would be 2 (the remainder left on dividing 5 by 3).

n = 1k2,k24
1+2+2+4=9
So, for n to be divis by 3, k must either be 0 or a multiple of 3 between 0 and 9. Therefore, k can be 0, 3, 6, or 9. Answer C.

I think the trick would be to consider '0' as potential candidate.

Hence four.

We know the following:

(1 + 2 + 2 + 4 + 2k)/3 = integer

(9 + 2k)/3 = integer

Thus, k must be a multiple of 3, so it can be 0, 3, 6, or 9.

Answer: C

Digits : 0,1,2,3,4,5,6,7,8,9

We can try every digit and see which fits perfect.

Only 0,3,6, and 9 works.

Hence n can have 4 values.

(C)

Am I the only one who found the question confusing? It asked "how many values could N have" - not K. I understand K can take 4 values, but doesn't any combination of 0, 3, 6, 9 mean a different value for N?