Missing Digit
What digit must replace the star to make the number a multiple of 11?
The 8-digit number 1234_678 is a multiple of 11.
What is the missing digit?
If you liked this problem, here is an NRICH task which challenges you to use similar mathematical ideas.
Answer: 9 (so the number is 12349678)
Using division
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![Missing Digit Missing Digit](/sites/default/files/styles/large/public/thumbnails/content-id-4990-missing%252520digit%252520from%252520beg.png?itok=xUh2KM1U)
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![Missing Digit Missing Digit](/sites/default/files/styles/large/public/thumbnails/content-id-4990-missing%252520digit%252520from%252520end%2525201.png?itok=OOAVFeB_)
The number is divisible by 11 so there must have been a remainder of 8 to give 88
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![Missing Digit Missing Digit](/sites/default/files/styles/large/public/thumbnails/content-id-4990-missing%252520digit%252520from%252520end%2525202.png?itok=qdOE5ZrM)
___7 gives a remainder of 8
107 = 99 + 8
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![Missing Digit Missing Digit](/sites/default/files/styles/large/public/thumbnails/content-id-4990-missing%252520digit%252520from%252520beg%252520and%252520end.png?itok=RAVAEz8K)
___6 gives a remainder of 10
76 = 66 + 10
2__ gives a remainder of 7
22 + 7 = 29
9 is missing
Using a divisibility test
A test for divisibility by 11 is to add alternate digits:
2__ gives a remainder of 7
22 + 7 = 29
9 is missing
Using a divisibility test
A test for divisibility by 11 is to add alternate digits:
1 + 3 + * + 7 = 11 + *; 2 + 4 + 6 + 8 = 20.
If the original number is a multiple of 11 then these two totals will be the same or will differ by a multiple of 11. In this case, 11 + * = 20 gives * = 9.
Using place value and algebra
1234*678 = 12340678 + 1000* = (11 x 1121879 +9) + 11 x 90* + 10*
and hence is divisible by 11 if and only if 10* + 9 is divisible by 11. So * = 9.