CRYPT multiplication 3

S

Saumyojit

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A=1

AARO= 11RO ---> (O *V + carry ) we can get 11 by 5*2+1 or 3*2+5 or 2*4 +3

G*i = unit digit will be A (1) . So, 3*7 or vice versa will be G and i.
@JeffM @Dr.Peterson
 
Saumyojit said:
View attachment 27868
A=1
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Showing an answer but no work is somewhat rude because determining whether your answer is correct is more difficult than necessary. It is also inefficient because explaining where you made an error if you made one is impossible.

But a = 1 is correct because t * a = t means t = 0 or a = 1. If t = 0, then t * i would equal t rather than o. Good work, but rude presentation.
AARO= 11RO ---> (O *V + carry ) we can get 11 by 5*2+1 or 3*2+5 or 2*4 +3

G*i = unit digit will be A (1) . So, 3*7 or vice versa will be G and i.
Click to expand...
I agree

[MATH]o(100v + 10i + 1) = 100ov + 10oi + o = 11(100) + 10r + 0 \implies oi = r + \alpha, ov + \alpha = 11 \implies\\ ov = 10 \text { and } \alpha = 1 \text { or }\\ ov = 8 \text { and } \alpha = 3 \text { or }\\ ov = 6 \text { and } \alpha = 5.[/MATH]It might have been helpful to say why you excluded 9 and carry 2, 7 and 4, 5 and 6, 4 +and 7, 3 and 8, and 2 and 9, but your result seems correct. It may be helpful later, but it leaves 6 possibilities open so it makes sense to explore some more.

I also agree gi = 21. Again you could give a little more information about that.

[MATH]gi = 10\beta + 1[/MATH] means both g and i are odd. Neither are 1 (because a = 1). Neither are 5. Both cannot be 9. So
g = 3 and i = 7 or else g = 7 and i = 3. Very nice. But notice that beta = 2.

[MATH]gv + 2 = 10a + i = 10 + i \implies gv = i + 8 \le 17.[/MATH]
What next?
 
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JeffM said:
It might have been helpful to say why you excluded 9 and carry 2, 7 and 4, 5 and 6, 4 +and 7, 3 and 8, and 2 and 9, but your result seems correct.
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what other possibilties?
O *V + carry =11 so if O =9 then 9*1+2=11 which is not possible as 1 is already taken.

JeffM said:
What next?
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Nothing is coming to my mind . I have been stuck in this place for past 5 hr
 
Saumyojit said:
what other possibilties?
O *V + carry =11 so if O =9 then 9*1+2=11 which is not possible as 1 is already taken.
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You are correct, but there is 3 * 3 + 2, not possible. Similar logic applies to the others. You gave ABSOLUTELY NO indication of your thinking and forced me to consider the other possibilities. If you had thought about other possibilities and rejected them, you did not show me the courtesy of sparing my time. I am a volunteer, do not get paid, and have other demands on my time. Why should I waste time on you when you do not show me even basic courtesy?
Nothing is coming to my mind . I have been stuck in this place for past 5 hr
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This is simply not true unless you ignored thinking about my post, which was submitted less than two hours ago.

You deduced that g = 7 and i = 3 or g = 3 and i = 7. What you failed to note is that the carry digit, which I called beta, is necessarily 2.

[MATH]\therefore gv + 2 = 10a + i = 10 + i \implies gv = i + 8 \le 17.[/MATH]
You have only two choices for g. Try g = 7. What then can you say about v? That means i = what?

How does that help you?
 
[MATH]gv = i + 8 \le 17 \implies 7v \le 17 \implies v = 2 \implies 7v = 14 = i + 8 \implies i = 6 \text { impossible.}[/MATH]
Therefore there is no need to assume g = 3 and i = 7. They must.

Glad you completed it on your own.
 
If anyone is interested in this as a puzzle, the solution is

a = 1, g = 3, v = 5, t = 6, i = 7, and s = 8.

Proof.

VIA = 571 and GOT = 326

6 * 571 = 3426 = GROT. Correct.

2 * 571 = 1142 = AARO. Correct.

3 * 571 = 1713 = AIAG. Correct.

3426 + 11420 + 171300 = 14,846 + 171,300 = 186,146 = ASTART = 571 * 326. Correct.
 
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