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What is the least number which must be subtracted from 1936 so that the remainder divided by 9,10,15 will leave in each case the same remainder 7?
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Understanding the Problem
To solve the problem, we need to find the least number that must be subtracted from 1936 so that the result leaves a remainder of 7 when divided by 9, 10, and 15.
Step 1: Setting Up the Equations
Let the number to be subtracted be x. We want:
- (1936 - x) % 9 = 7
- (1936 - x) % 10 = 7
- (1936 - x) % 15 = 7
This means:
- 1936 - x = 7 (mod 9)
- 1936 - x = 7 (mod 10)
- 1936 - x = 7 (mod 15)
Step 2: Finding the Remainders
We can rewrite the equations as:
- 1936 - 7 = x (mod 9)
- 1936 - 7 = x (mod 10)
- 1936 - 7 = x (mod 15)
Calculating 1936 - 7:
- 1936 - 7 = 1929
Now, we need to find x such that:
- 1929 % 9
- 1929 % 10
- 1929 % 15
Step 3: Calculate Remainders
- 1929 % 9 = 6
- 1929 % 10 = 9
- 1929 % 15 = 9
Step 4: Calculating x
To achieve a remainder of 7, we need:
- For mod 9: We need 1929 to be adjusted to 7. Thus, we subtract 6 from 1929.
- For mod 10: We need to adjust to 7, thus subtracting 2.
- For mod 15: We need to adjust to 7, thus subtracting 2.
The highest adjustment required is 6.
Conclusion
Therefore, the least number that must be subtracted from 1936 is:
- x = 6
Thus, the answer is 6, ensuring that the remainder when dividing by 9, 10, and 15 is consistently 7.
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What is the least number which must be subtracted from 1936 so that the remainder divided by 9,10,15 will leave in each case the same remainder 7?
Question Description
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