What is the sum of the digits of the ID number of my ATM card?
What is the sum of the digits of the ID number of my ATM card?
The correct option is B 45
The alphabets PQRSTUVWX represent the nine-digit number with the given property.
Deduction 1 : Since all numbers having even digits have to be divisible by even numbers (i.e., PQ is divisible by 2, PQRS is divisible by 4, PQRSTU is divisible by 6 and PQRSTUVW is divisible by 8), Q, S, U & W must be the 4 even numbers viz. 2, 4, 6 and 8 in some random order. Consequently, P, R, T, V & X (the odd placed digits) must be sharing the odd digits 1,3, 5,7 and 9 in some order.
Deduction 2: Since PQRST is divisible by 5, T must be equal to 5. Thus, the number must be PQRS5UVWX.
Deduction 3: PQR is divisible by 3 and PQRSTU is divisible by 6, hence STU must be divisible by 3, i.e., S + T+U=S+5+U must be divisible by 3. Also, S and U are even numbers. Through trial and error you should realise that there are only two ways this could happen:
(a) If STU represents 258
or (b) If STU represents 654
Note: S can only be 2 or 6 since for PQRS to be divisible by 4, RS should be divisible by 4 and R being an odd number, S can only be 2 or 6.
Deduction 4: The following 4 options of filling in the other two even numbers (4 and 6 in case STU is 258 or 2 and 8 in case STU is 654) emerge:
PQRSTUVWXPossibility 1 4 258 6 Possibility 2 6 258 4 Possibility 3 2 654 8 Possibility 4 8 654 2
Sum of digits - 1+2+3+4+5+6+7+8+9=45
Hence (b).
45
The alphabets PQRSTUVWX represent the nine-digit number with the given property.
Deduction 1 : Since all numbers having even digits have to be divisible by even numbers (i.e., PQ is divisible by 2, PQRS is divisible by 4, PQRSTU is divisible by 6 and PQRSTUVW is divisible by 8), Q, S, U & W must be the 4 even numbers viz. 2, 4, 6 and 8 in some random order. Consequently, P, R, T, V & X (the odd placed digits) must be sharing the odd digits 1,3, 5,7 and 9 in some order.
Deduction 2: Since PQRST is divisible by 5, T must be equal to 5. Thus, the number must be PQRS5UVWX.
Deduction 3: PQR is divisible by 3 and PQRSTU is divisible by 6, hence STU must be divisible by 3, i.e., S + T+U=S+5+U must be divisible by 3. Also, S and U are even numbers. Through trial and error you should realise that there are only two ways this could happen:
(a) If STU represents 258
or (b) If STU represents 654
Note: S can only be 2 or 6 since for PQRS to be divisible by 4, RS should be divisible by 4 and R being an odd number, S can only be 2 or 6.
Deduction 4: The following 4 options of filling in the other two even numbers (4 and 6 in case STU is 258 or 2 and 8 in case STU is 654) emerge:
PQRSTUVWXPossibility 1 4 258 6 Possibility 2 6 258 4 Possibility 3 2 654 8 Possibility 4 8 654 2
Sum of digits - 1+2+3+4+5+6+7+8+9=45
Hence (b).
