Playing with Numbers Test - 4

LCM of 6, 8 and 9 is 72.
Required number = 72x + 5

Put x = 6 and x = 7 to get the numbers between 400 and 550.
Required numbers = 72 × 6 + 5, 72 × 7 + 5 = 437 and 509

57w88 is divisible by 18 if the digit at the units place is even and the sum of the digits is divisible by 9.
5 + 7 + w + 8 + 8 = 28 + w is divisible by 9.

The multiples of 9 are 9, 18, 27, 36 ...
Next multiple of 9 after 27 is 36, so we need to add 8 to make it 36.

So, least possible value of w = 8

A number is said to be divisible by 6 if it is divisible by 2 and 3.

Option 2 is correct here because dfg is an even number or it is divisible by 2. Also, the sum of the digits in the number is divisible 3.
E.g. 756

756 is an even number.
Also, 7 + 5 + 6 = 18, which is divisible by 3.

So, 756 is divisible by 6.

According to the question,
Dividend = Divisor × Quotient + Remainder

Let the quotient be Q and divisor be D.
So, dividend is 12Q.
Also, remainder will (1/6) times the dividend i.e. 2Q.

Therefore,
Dividend = Divisor × Quotient + Remainder
12Q = D × Q + 2Q
10Q = DQ
D = 10

(A) If the last two digits of a whole number are divisible by 4, then the entire number is divisible by 4.
E.g. 4316; 16 is divisible by 4, which means 4316 is divisible by 4.
So, this statement is true.

(B) If the sum of the digits is divisible by 9, then the actual number is also divisible by 9.
E.g. 1233; sum of the digits = 1 + 2 + 3 + 3 = 9 (which is divisible by 9). So, 1233 is also divisible by 9.
So, this statement is true.

(C) If the last three digits of a whole number are divisible by 7, then the entire number is divisible by 7.
E.g. 1133; 133 is divisible by 7, but 1133 is not divisible by 7.
So, this statement is not true.

(D) Take the difference between alternating sum of the digits in the number. If that is divisible by 11, so is the original number.
For instance, 7282 has the alternating sum of digits as 7 - 2 + 8 - 2 = 11. Since 11 is divisible by 11, 7282 is divisible by 11.
So, this statement is true.

(i) If 42a5 is a multiple of 9 and 'a' is a digit, then the sum of the digits is: 4 + 2 + a + 5 = 11 + a, which should be divisible by 9.
So, the value of 'a' will be 7, as 11 + 7 = 18, which is divisible by 9.
(ii) The usual form of the number 9 × 100 + 7 × 1 is: 900 + 7 = 907.
(iii) It is given that M is a number such that M ÷ 5 gives a remainder of 1.

Since every multiple of 5 ends at 0 or 5, to get the remainder of 1, the ones digit should be one more than 0 or 5. So, the ones digit must be 1 or 6.

For a number to be divisible by 9, the sum of the digits should be divisible by 9.
That is, 3 + 7 + 4 + a = 14 + a = 14 + 4 = 18, which is divisible by 9.
(A) 3474 ÷ 9 = 386

(B) Sum of the digits = 2 + 8 + 2 + 2 + 1 = 15, which is divisible by 3.
Hence, 28,221 ÷ 3 = 9407

(C) If a number is divisible by 5, then the digit on the ones place can be either 0 or 5.
Hence, the correct answer is 5.

(D) For a number to be divisible by 11, take the alternating sum of the digits in the number, read from left to right. If that is divisible by 11, so is the original number. So, 2X28 has the alternating sum of digits as (2 + 2) - (X + 8) = a number that is divisible by 11.

Therefore, the least number can by 7 as (2 + 2) - (7 + 8) = 4 - 15 = -11, which is divisible by 11.
Hence, 2 X = 2 × 7 = 14.

If the sum of the digits is a multiple of 9 then, the number is divisible by 9.
Hence, if the number is AABAA, then 4A + B must be a multiple of 9.

For any value of A, B can be adjusted to make 4A + B a multiple of 9.
Therefore, there will be 9 such numbers as the value of A will vary from 1 to 9.

If A = 1, then B = 5
If A = 2, then B = 1
If A = 3, then B = 6
If A = 4, then B = 2
If A = 5, then B = 7
If A = 6, then B = 3
If A = 7, then B = 8
If A = 8, then B = 4
If A = 9, then B = 0