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Missing digits (Posted on 2011-06-06) Difficulty: 3 of 5
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

No Solution Yet Submitted by Ady TZIDON    
Rating: 3.6667 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution mostly computer solution (spoilers) | Comment 1 of 8

The first one is easy enough (D1 it says) to get by hand. The digits already given add up to 19 and must be supplemented to get to a multiple of 9 (like 27), and be such that the last digit is 0 or 5. If the last digit is 0, the first ? must be replaced by 8. If the last digit is 5, the first ? must be replaced by a 3.

The computer solution agrees with that double solution for the first problem.  All the others have just one solution each, the third one also calling for two answers: the number and the quotient when divided by 417.

3435615
3485610

689337

9141057  21921

7054344

49187

6780187951

The Frink program:

for a=0 to 9
{
for b=0 to 5 step 5
 {
 n=3405610 + a*10000+b
 if n%45==0
    println[n]
 }
}
println[]

for a=0 to 99
{
 n=680037 + a*100
 if n%99==0
    println[n]
}
println[]


for a=0 to 999
{
 n=9000057+a*1000
 if n%417 ==0
  {
  q=n/417
  tst=int[q/1000]%10
  if tst==1
   println["$n  $q"]
  }
}
println[]

for a=0 to 99
{
for b=0 to 8 step 2
 {
 n=7000340 + a*1000+b
 if n%792==0
    println[n]
 }
}
println[]

for a=0 to 9
{
for b=0 to 9
 {
 n=40180 + a*1000+b
 if n%101==0
    println[n]
 }
}
println[]

for a=0 to 9
{
for b=0 to 9
for c=0 to 99
 {
 n=6080080051 + a*100000000+b*100000+c*100
 if n%(73*137)==0
    println[n]
 }
}
println[]


  Posted by Charlie on 2011-06-06 14:17:37
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