Quantitative Ability Question of the Day

Question:-

How many 4-digit numbers divisible by 7 have at least two zeroes?

OPTIONS

1) 34

2) 32          
3) 30
4) 28
5) 37

Solution


The 4 digit numbers divisible by 7 and having at least two zeroes can be further divided into three sub cases, numbers of the form ab00, x0y0 and m00n.


Case 1: Numbers are of the type ab00
For ab00 to be divisible by 7, ab should be divisible by 7.
Clearly, ab will take values 14, 21, 28, 35, …, 98
So, there are 13 solutions in this case.
Case 2: Numbers are of the type x0y0
For x0y0 to be divisible by 7, 1000x + 10y should be divisible by 7.
1000x + 10y = 1001x – x + 7y + 3y
As 1001x + 7y is divisible by 7, 3y – x has to be divisible by 7.
∴3y − x = 0, ±7, ±14 or ±21.
For 3y − x = 0, possible solutions for (x, y) are (3, 1), (6, 2), (9, 3).
For 3y − x = 7, possible solutions for (x, y) are (2, 3), (5, 4) and (8, 5).
3y − x = –7 is not possible for positive x and y.
For 3y − x = 14, possible solutions for (x, y) are (1, 5), (4, 6) and (7, 7).
3y − x = –14 is not possible for positive x and y.
For 3y − x = 21, possible solutions for (x, y) are (3, 8) and (6, 9)
3y − x = –21 is not possible for positive x and y.

So, there are 11 solutions in this case.

Case 3: Numbers are of the type m00n
For m00n to be divisible by 7, 1000m + n should be divisible by 7.
∴ n − m should be divisible by 7.
∴ n − m = 0, ±7.
For n – m = 0, possible solutions for (m, n) are (1, 1), (2, 2),…, (9, 9)
For n − m = 7, possible solutions for (m, n) are (1, 8) and(2, 9).
For n − m = –7, possible solutions for (m, n) are (8, 1) and (9, 2).
So, there are a total of 13 solutions in this case.

∴ Total number of solutions = 13 + 11 + 13 = 37

Hence, option 5.