[Guest]

Substitute each of the capital letters in bold by a different base ten digit from 0 to 9 to satisfy this alphametic equation. None of the numbers can contain any leading zero.

(NOD)² = ONLOAD

Note: While a solution is trivial with the aid of a computer program, show how to derive it without one.

[Guest]

[spoiler='Answer

']NOD x NOD = onLOAD

NOD = onLOAD/NOD

NOD = LOA

ND = LA

Then, just plug in any numbers that you want to make this work.

First time posting not sure if I got it correct.

Edited June 24, 2009 by MattMat

[Pickett]

so, we know that NOD must be between 317 and 999 (inclusive) in order to get a 6 digit number

We also know that D must be either 0, 1, 5, or 6 so that D * D = a number ending in D...

However, if D were 0, then the last 2 digits of the result would be D...because ##0 * ##0 = ###00...so D can't be 0.

So given that, we know:

N must be 3, 4, 5, 6, 7, 8, or 9

D must be 1, 5, or 6

and O cannot be 0.

The basic pattern now is simply to try the values of N, and what you find is that when you choose a specific value for N, there are only 2 or 3 numbers you have to try out. Let me show by example:

If N was 3, then O must be 1 because 396 (highest number allowed with N=3) squared is 156816...But that means the only two numbers it could be are 315 and 316...which we've already shown are too low for it. So N cannot be 3.

If N was 4, then O must be 1 or 2...and if O is 2, then 426 (lowest number allowed with N=4, O=2) squared results in a number beginning with 1...so O can't be 2, O must be 1...but once again, we run into the problem that 415 squared (smallest number allowed with this pattern) is too high to result in a 14#### number. So N can't be 4.

If N was 5, then O must be 2 or 3...if O is 2, our lowest number is 521, which when squared gives a number higher than 25####, so O can't be 2...so O must be 3...but, 536 squared ends up still only yielding 2#####, so O can't be 3, and N can't be 5...

If N was 6, then O must be 3 or 4 (catch the pattern here so far)? If O is 3, then 631 is the only number to yield a value 3#####, but unfortunately it doesn't fit the pattern, so O can't be 3. O must be 4...but again, we find that 641 and 645 do not fit...so N can't be 6

If N was 7, then O must be 5 or 6. If O is 5, then we check 751 and 756 as our possible values...and lo and behold, we find that 756 works:

(NOD)² = onLOAD

(756)² = 571536

So

N = 7

O = 5

D = 6

L = 1

A = 3

Keep 'em coming K, whenever I don't have work to be doing, these are a great way to spend my time

[Guest]

divide both sides by NOD

so u get NOD=LOA

then it realy doesnt matter what O is so ill make it 1

so ND=LA

2x10=20 and 4x5=20

so N=2 D=10 L=4 A=5

[Guest]

"Keep 'em coming K, whenever I don't have work to be doing, these are a great way to spend my time"

Definately keep these up, I find these alphanumeric puzzles very entertaining.

Seems Pickett got to this before I did. Hopefulyl I catch the next one first.

[Guest]

divide both sides by NOD

so u get NOD=LOA

then it realy doesnt matter what O is so ill make it 1

so ND=LA

2x10=20 and 4x5=20

so N=2 D=10 L=4 A=5

Nice try, crabs, but each letter is a digit, and so D cannot equal 10

[Guest]

just a heads up to anyone dividing by a letter, not only does it not work because they are digits not numbers, but even if they were numbers, you have to be careful with one of them possibly being zero. As dividing by zero is a mathematical cardinal sin...or atleast an impossibility

*learned that one in junior high algebra when my teacher had "mathematically proven" that 1=2

Edited June 25, 2009 by perk8504

[Guest]

[spoiler='Answer

']NOD x NOD = onLOAD

NOD = onLOAD/NOD

NOD = LOA

ND = LA

Then, just plug in any numbers that you want to make this work.

First time posting not sure if I got it correct.

Nice try. But that was an answer to a different problem.

This is an alphametic puzzle in conformity with the definition given here.

[Guest]

so, we know that NOD must be between 317 and 999 (inclusive) in order to get a 6 digit number

We also know that D must be either 0, 1, 5, or 6 so that D * D = a number ending in D...

However, if D were 0, then the last 2 digits of the result would be D...because ##0 * ##0 = ###00...so D can't be 0.

So given that, we know:

N must be 3, 4, 5, 6, 7, 8, or 9

D must be 1, 5, or 6

and O cannot be 0.

The basic pattern now is simply to try the values of N, and what you find is that when you choose a specific value for N, there are only 2 or 3 numbers you have to try out. Let me show by example:

If N was 3, then O must be 1 because 396 (highest number allowed with N=3) squared is 156816...But that means the only two numbers it could be are 315 and 316...which we've already shown are too low for it. So N cannot be 3.

If N was 4, then O must be 1 or 2...and if O is 2, then 426 (lowest number allowed with N=4, O=2) squared results in a number beginning with 1...so O can't be 2, O must be 1...but once again, we run into the problem that 415 squared (smallest number allowed with this pattern) is too high to result in a 14#### number. So N can't be 4.

If N was 5, then O must be 2 or 3...if O is 2, our lowest number is 521, which when squared gives a number higher than 25####, so O can't be 2...so O must be 3...but, 536 squared ends up still only yielding 2#####, so O can't be 3, and N can't be 5...

If N was 6, then O must be 3 or 4 (catch the pattern here so far)? If O is 3, then 631 is the only number to yield a value 3#####, but unfortunately it doesn't fit the pattern, so O can't be 3. O must be 4...but again, we find that 641 and 645 do not fit...so N can't be 6

If N was 7, then O must be 5 or 6. If O is 5, then we check 751 and 756 as our possible values...and lo and behold, we find that 756 works:

(NOD)² = onLOAD

(756)² = 571536

So

N = 7

O = 5

D = 6

L = 1

A = 3

Keep 'em coming K, whenever I don't have work to be doing, these are a great way to spend my time

Well done, and a great solution yet once more !!!!