# Flock of Geese alphametic

In each of the following three alphametic puzzles, every letter stands for a digit in base-10 representation, and different letters stand for different digits. Leading digits are always non-zero.

   FLOCK * 6 = GEESE

FLOCK * 7 = GEESE

FLOCK * 8 = GEESE


Which digit does each letter represent? (Please present the full analysis how these digits can be determined.)

• Can it be assumed that the F is not zero? Ot should I not rule that out?
– Ivo
Commented Feb 14, 2016 at 20:04
• It's not stated if the number value associated with a letter can change between cases. This would make a huge difference. Commented Mar 4, 2016 at 7:44

# PART 2

For FLOCK * 7 = GEESE, the answer is:

F=1, L=0, O=9, C=4, K=8, G=7, E=6, S=3

Explanation:

F can only be 1 to maintain the number of digits, therefore, G=7/8/9 and L=0/2/3/4

Since 7 is odd, we can produce numbers with patterns 700#0, 711#1, 722#2, etc. until 988#8.

I started with 700#0. We can establish 70000 and 70070 are divisible by 7. If we keep adding either 70 or 1071 (which is 7 * 53) as needed, we can establish the pattern:

70000, 70070, 71141, 72212, 72282, 73353, 74424, 74494, 75565, 76636, 77707, 77777, 78848, 79919, 79989

Finding the next number divisible by 7 with the format 800#0 and continuing the pattern, we get:

80010, 80080, 81151, 82222, 82292, 83363, 84434, 85505, 85575, 86646, 87717, 87787, 88858, 89929, 89999

With 900#0, we get

90020, 90090, 91161, 92232, 93303, 93373, 94444, 95515, 95585, 96656, 97727, 97797, 98868, 99939

Removing the ones with:

• 1s on any digit,
• repeated tens digit,
• repeated ten-thousands digit,
• 0s formatted 7####,
• 2s formatted 8####, and
• 3s or 4s formatted 9####

We get:

72282, 73353, 74424, 74494, 75565, 76636, 78848, 79989, 80080, 82292, 83363, 84434, 85505, 85575, 86646, 89929, 90020, 90090, 95585, 96656, 97727, 98868

I tried dividing it all by 7 and the only working solution is

10948 * 7 = 76636

## Part 3

Assuming F is at least 1, FL must be either 12 or 10 for the product to be less than 10000.

If FL is 12, then GEESE is at least 96000. E must be even, so it can only be 6 or 8. If E is 6, then FLOCK*8 being between 96606 and 96686 requires that O is 0 and C is 8. K can't be 2, so it must be 7 to make the units digit correct. But 12087*8=96696 duplicates the digit 9, so E being 6 doesn't work. E must be 8, making O be 3 and C be 5 or 6. K can't be 1, so it is 6 and C is 5. The result 12356*8=98848 works.

If FL is 10, then GEESE is at least 80000 and at most 87999. Again, E must be even, so it has to be 2, 4, or 6. If E is 2, then O must also be 2, which is impossible. If E is 4, then O must be 5, and C must be 6 to avoid a repeat. Then GEESE is either 84484 or 84494, but neither of these numbers is divisible by 8. Finally, if E is 6, then O must be 8, which is already used as G. So none of these cases works.

The only working answer is 12356*8=98848.

# Partial answer (only first puzzle)

My analysis:
F is 1 (assuming it can't be 0) because otherwise the result would be 6 digits, in all cases.

For the first case:
GEESE is divisible by 6 which means that it's even, so E must be an even number.
GEESE is divisible by 6 which means it's also divisble by 3 which means that the digit sum must be divisible by three which means that G+S is divisble by three.
Because G is at least 6 times F then G is 6 or higher

Now by these conclusions the possibilities for GEESE is a list that I thought was small enough to check by hand:
60030 60090 62232 62292 64434 64494 68838 68898 70020 70050 70080 72252 72282 74424 74454 74484 76626 76656 76686 78828 78858 80040 80070 82242 82272 84474 86646 86676 90030 90060 92202 92232 92262 94404 94434 94464 96606 96636 98808 98838 98868

So I checked every one of them what they were divided by 6 and if they would fit the equation. My result:

The solution is $15367 * 6 = 92202$

• In these puzzles I think the first digit can't ever be a 0, so for F=1 this should be alright. Commented Feb 14, 2016 at 21:02