Question

In a 3-digit number, sum of the digits is 15. One of the digits is 5 more than the second
digit and second digit is 2 more than the third digit. How many solutions are possible?​

Answer 1

Answer:

Let the 3 digit number be XYZ.

According to given conditions.

Y = 5X

Z = Y -4

As X,Y,Z are digits

0≤ X,Y,Z ≤ 9

Now

0≤ Z≤9

0≤ Y-4 ≤ 9

4≤ Y ≤ 13

But

0≤ Y ≤ 9

Common part is

4≤ Y ≤ 9

So,

0≤ Z ≤ 5

Now , Y = 5X .

4≤ 5X ≤ 9

4/5≤ X ≤ 9/5.

0.8≤ X ≤ 1.8

As X is a digit it should be a non negative integer. The only integer between 0.8 and 1.8 is 1 .

So X = 1.

Then ,Y = 5X = 5.

Therefore Z = Y-4 = 1

Thus the number is 151.

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