### Brain Teasers

# The Troll Bridge 6

A bridge was guarded by an evil troll. The troll was very intelligent, but he was also a coward. He was afraid of anyone smarter than him. So every time anyone tried to cross the bridge, the troll would set up a test. If the traveler passed the test, he would be allowed to cross. Otherwise, the troll would eat him.

A traveler came across the bridge. The troll said, "You may only cross my bridge if you know the password." He then wrote an equation on a rock:

SLIDE

-DEAN

______

3651

"Each letter in the equation represents a different digit," said the troll. "The letters in the password represent 3651."

So, what is the password?

A traveler came across the bridge. The troll said, "You may only cross my bridge if you know the password." He then wrote an equation on a rock:

SLIDE

-DEAN

______

3651

"Each letter in the equation represents a different digit," said the troll. "The letters in the password represent 3651."

So, what is the password?

### Answer

The password is LENS.Since SL-D=3 or (SL-1)-D=3 (meaning SL-D=4), and since the highest possible sum of two different one-digit numbers is 17, S must equal 1.

Since SL=1L=10+L, D must be 6 or greater, since 6 is the lowest number that can be added to 4 to get a two-digit number.

Suppose D=6. Then L must equal 0. Since D-A (or 1D-A, but this is impossible because 16 or greater minus 5 or 6 is a two-digit number) is 5 or 6, A must equal 0 or 1. But since L=0 and S=1, this cannot work.

Suppose D=7. Then L is 0 or 1, and since S=1, L=0. A must be 1 or 2, but since S=1, A=2.

Suppose D=8. Then L is 1 or 2, and since S=1, L=2. A must be 2 or 3, but since L=2, A=3.

Suppose D=9. Then L is 2 or 3, and A is 3 or 4. Therefore, either L=2 and A=3, L=2 and A=4, or L=3 and A=4.

The only possibility for D-A=6 is D=9 and A=3. Since 1 is carried from D to E, N must equal 9, since that's the only one-digit number that can be added to 1 to get a two-digit number. But since D=9, this cannot work.

Therefore, D-A=5, and E-N=1. Since 1 could not have been carried from D to E, E cannot be 0. Since S=1, E cannot be 1. If E=2, then N=1, which cannot work because S=1. If E=3, then N=2. But if D=7, A=2. If D=8, then L=2 and A=3. If D=9, then L is either 2 or 3. So E cannot be 3.

Therefore, E is 4 or greater. Since 4+6 is a two-digit number, 1 must have been carried over from L to I, and 1L-D=4.

If D=7, then L=0. But since 10-7=3, D cannot be 7.

If D=9, then L=3 and A=4.

If E=4, then N=3. But if D=8, A=3, and if D=9, L=3 and A=4. So this cannot work.

If E=5, then since 1I-E=6, I=1. But since S=1, this cannot work.

If E=7, then I=3. But if D=8, then A=3, and if D=9, then L=3. So this cannot work.

If E=8, then I=4, and D cannot be 8. But if D=9, then A=4. So this cannot work.

If E=9, then N=8, and D cannot be 8 or 9. So this cannot work.

Therefore, E=6, N=5, and I=2. If D=8, then L=2, so D must equal 9. This means:

S=1, I=2, L=3, A=4, N=5, E=6, and D=9.

Therefore, 3651=LENS.

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