Question

12
Identify me.
O I am a multiple of 7. Bigger than 40 and smaller than 50. The sum of my two digit
is 13. Identify me?​

Answer 1

Question :- I am a multiple of 7. Bigger than 40 and smaller than 50. The sum of my two digit is 13. Identify me ?

Solution :-

Let the required two digit number is x .

given that,

• x is multiple of 7 => x ÷ 7 = Remainder 0 .
• greater than 40 and smaller than 50 => 7 * 6 = 42 or 7 * 7 = 49 .
• sum of digits is 13.

so,

→ x = 42 or 49 .

checking sum of digits we get,

• if x = 42 => sum of digits = 4 + 2 = 6 .
• if x = 49 => sum of digits = 4 + 9 = 13.

therefore, we can conclude that the required number is 49.

Learn more :-

if a nine digit number 260A4B596 is divisible by 33, Then find the number of possible values of A.

https://brainly.in/question/32686002

if n is an integer such that 1nn352 is a six digit number

https://brainly.in/question/26617043

Answer 2

Given:

I am a multiple of 7. Bigger than 40 and smaller than 50. The sum of my

two-digit  is 13.

To find:

The required number

Solution:

It is given that the number is a multiple of 7, it is bigger than 40 and smaller than 50.

The multiples of 7 bigger than 40 and smaller than 50.

=> 7 x 6 = 42

=> 7 x 7 = 49

So, 42 and 49 are the multiples bigger than 40 and smaller than 50

It is given that the sum of its digits are 13

=> 4 + 2 = 6

=> 4 + 9 = 13

49 is a multiple of 7, Bigger than 40 and smaller than 50, the sum of its digits are 13.

Hence, the required answer is 49.