What least number must be subtracted from 1936 so that the resulting number when divided by 9, 10 and 15 will leave in each case the same remainder 7?

Options

A. a) 37
B. b) 36
C. c) 39
D. d) 30

Answer

To find the least number that must be subtracted from 1936 so that the resulting number leaves a remainder of 7 when divided by 9, 10, and 15, we can follow these steps:

### Step 1: Find the LCM of 9, 10, and 15
To ensure that the resulting number leaves the same remainder when divided by 9, 10, and 15, we first need to find the least common multiple (LCM) of these three numbers.

- The prime factorization of each number is:
- 9 = 3²
- 10 = 2 × 5
- 15 = 3 × 5

- The LCM is found by taking the highest power of each prime:
- For 2: highest power is 2¹ from 10
- For 3: highest power is 3² from 9
- For 5: highest power is 5¹ from both 10 and 15

Thus, the LCM = 2¹ × 3² × 5¹ = 2 × 9 × 5 = 90.

### Step 2: Determine the target number
We want the resulting number (1936 - x) to leave a remainder of 7 when divided by 9, 10, and 15. This means we can express the target number as:
\[ 1936 - x \equiv 7 \mod 90 \]
This can be rewritten as:
\[ 1936 - x = 90k + 7 \]
for some integer \( k \).

### Step 3: Rearranging the equation
Rearranging gives us:
\[ x = 1936 - (90k + 7) \]
or
\[ x = 1929 - 90k \]

### Step 4: Find the least value of x
To find the least value of \( x \), we need to maximize \( k \) such that \( x \) remains non-negative:
\[ 1929 - 90k \geq 0 \]
This simplifies to:
\[ k \leq \frac{1929}{90} \approx 21.6 \]
Thus, the maximum integer value for \( k \) is 21.

### Step 5: Calculate x for k = 21
Substituting \( k = 21 \):
\[ x = 1929 - 90 \times 21 \]
Calculating this gives:
\[ x = 1929 - 1890 = 39 \]

### Conclusion
The least number that must be subtracted from 1936 is **39**.

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