1 × 2 × ⋯ × 1 5 = 1 3 0 7 6 7 a 3 6 8 0 0 0
What is the missing digit a ?
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"Note: If the digit sum without is a multiple of 9, then could only be 0 or 9. We will need to do more work."
This gives me an inspiration !
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I like that question!
Eric's approach would work better for this problem, as we will always get just 1 answer. If more digits were hidden, then we will need more tests.
That's essentially the logic.
Can you complete the calculation to explain how to arrive at the answer? Thanks!
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Thanks. Great explanations and tips that you pointed out.
I've edited the presentation order so that it's easier to read the solution.
1 5 ! has ⌊ 2 1 5 ⌋ + ⌊ 2 2 1 5 ⌋ + ⌊ 2 3 1 5 ⌋ = 7 + 3 + 1 = 1 1 factors of 2 . You can be sure that the last n digits is divisible by 2 n ; for example a 3 6 8 0 0 0 is divisible by 2 7 or a 3 6 8 is divisible by 2 4 , if the double possibility of 0 or 9 arises.
Similar to Saya Suka's solution, but with 11.
Let each digits in 130767a368000 have indexes, where the first digit (1) has an index of 1, the second digit (3) has an index of 2, and etc.
The divisibility test for 11 is that the sum of all the digits with even indexes differ from the sum of all the digits with odd indexes by a multiple of 11 (including 0).
Sum of digits with odd digits: 1 + 0 + 6 + a + 6 + 0 + 0 = 13 + a Sum of digits with even digits: 3 + 7 + 7 + 3 + 8 + 0 = 28
13 + a = 28 + 11n
a= 15 + 11n
Since a is a digit, 0<= a <= 9.
Thus, n = -1 and a = 4
Nice! The benefit of using the rule of 11 is that we're guaranteed a unique answer no matter which digit is hidden (assuming that all the other digits are correct).
11 might be a bit overkill for this problem, but is the best method in this variation !
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I don't think it's an overkill because of how simple the rule is. If we tried to apply divisibility by 17 which doesn't have a real simple use, then that would be an overkill.
Same way :) (+1)
Find sum of all digits except a
thats 41 but it should be a multiple of 3... so 4 is the only possible option
Also, 1 and 7 make the number divisible by 3 .
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The divisibility test for 9 is applicable here (which is glaringly obvious since 1 ≤ 9 ≤ 1 5 ).
The total sum of the digits will be equal to some multiple of 9.
We have 1 × ( 1 + 8 + a ) + 2 × ( 3 + 6 + 7 ) + 4 × 0 is a multiple of 9.
This is equal to 4 1 + a .
Since 0 ≤ a ≤ 9 , hence we must have a = 4 to give 4 1 + 4 = 4 5 as the multiple of 9.
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