Given:
- Number to be subtracted from 1936 to get a number that leaves the same remainder when divided by 9, 10, and 15 is to be found.
- Remainder when divided by 9, 10, and 15 is 7.

To find:
- The number to be subtracted.

Solution:

Step 1: Finding the LCM of 9, 10, and 15.

- LCM of 9, 10, and 15 = 2² × 3² × 5 = 180.

Step 2: Finding the number that leaves a remainder of 7 when divided by 9, 10, and 15.

- We can add 7 to multiples of 180 to get the required number.
- First, we find the smallest multiple of 180 greater than 1936:

1936 ÷ 180 = 10 remainder 36

Next multiple of 180 = (10+1) × 180 = 1980

- Now, we add 7 to 1980:

1980 + 7 = 1987

Therefore, 1987 is the smallest number that leaves a remainder of 7 when divided by 9, 10, and 15.

Step 3: Finding the number to be subtracted.

- We need to subtract a number from 1936 such that the resulting number leaves the same remainder of 7 when divided by 9, 10, and 15.

- Let the number to be subtracted be x.

(1936 - x) ÷ 9 leaves a remainder of 7

(1936 - x) ÷ 10 leaves a remainder of 7

(1936 - x) ÷ 15 leaves a remainder of 7

- We can write this as:

(1936 - x) = 7 + 9k (where k is an integer)

(1936 - x) = 7 + 10m (where m is an integer)

(1936 - x) = 7 + 15n (where n is an integer)

- Rearranging the equations, we get:

x = 1929 - 9k

x = 1926 - 10m

x = 1929 - 15n

- We need to find the smallest value of x that satisfies all three equations.

- We can substitute the values of k, m, and n in the equations and check which value of x satisfies all three equations.

For k = 0, x = 1929

For m = 0, x = 1926

For n = 0, x = 1929

- The smallest value of x that satisfies all three equations is 1929.

- Therefore, the number to be subtracted from 1936 is 1929 -

1936 - 1929 = 7

- Thus, the correct option is (c) 39.

Therefore, the least number that must be subtracted from 1936 so that the resulting number when divided by 9, 10, and 15 will leave the same remainder 7 is 39.