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 Missing digits (Posted on 2011-06-06)
Replace the interrogation marks in the following statements by appropriate digits (0 to 9) to make these statements valid:

D1: 34?561? is exactly divisible by 45
D2: 68??37 is exactly divisible by 99
D3: 9???057 is equal to 417 times ?1???
D3: 70??34? is exactly divisible by 792
D4: 4?18? is exactly divisible by 101
D5: 6?80?8??51 is exactly divisible by 73*137

Comments: ( Back to comment list | You must be logged in to post comments.)
 A few by hand | Comment 2 of 8 |
For 34?561? to be divisible by 45 = 5*9, its digits must sum to a multiple of 9 and it must end in 0 or 5.  The given digits sum to 19 so the unknown digits give 8 more.  There are two possibilities (8,0) and (3,5).  So the number is either 3485610 or 3435615.

For 68??37 to be divisible by 99 = 9*11 its digits must sum to a multiple of 9 and it has to meet the divisibility rule for 11.  The given digits sum to 24 so the unknown digits give 3 or 12 more.  The digits in the odd places sum to 6 more than the even places; this differential has to be brought down by 6 to 0 or up by 5 to 11.   Digits summing to 3 cant do this, digits summing to 12 can only differ by an even amount.  In this case 9-3=6 is what we want.  So the number is 689337.

4?18? is also very easy (D2) if you set up a long multiplication:
`  abcx 101  abcabc4?18?`

b must be 8, a+c carries to make the first ?=9, then a=4, so c=7 as is the second ?.  So the number is 49187

 Posted by Jer on 2011-06-06 15:07:01

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