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Crypt arithmetic multiplication 4
- Thread starter Saumyojit
- Start date
D
Deleted member 4993
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Strange ...... just a fleeting look tells meView attachment 27875
I tried but nothing came .there is nothing to Show
U + B < 10
It is so easy to say "there is nothing to Show"
These take a while. And they repay being systematic as I have said BEFORE.
I suggest that you write down a grid with the letters going across and the digits going down to keep track of what you find.
For w, n, and b, you can immediately put no in the respective zero cells.
Now you have the same trick as in the previous puzzle.
[MATH]u * y = y \implies u = 1 \text { or } y = 0.[/MATH]
[MATH]\text {If } y = 0 \implies t * y = 0 + 10 * 0 = y \ne p. \text { Thus, } n = 1 \text { and } y \ne 0.[/MATH]
So you can put yes in the cell for u = 1. Put no in all the other 1 cells, and put no in y’s zero cell.
Now there are two things that look a bit interesting, but I I do not know whether they lead anywhere.
One is that [MATH]t(100w + 10h + y) = 1100b + 10n + p.[/MATH]
The other is that [MATH]carry-digit + b + u = n \le 9.[/MATH] Why?
I suggest that you write down a grid with the letters going across and the digits going down to keep track of what you find.
For w, n, and b, you can immediately put no in the respective zero cells.
Now you have the same trick as in the previous puzzle.
[MATH]u * y = y \implies u = 1 \text { or } y = 0.[/MATH]
[MATH]\text {If } y = 0 \implies t * y = 0 + 10 * 0 = y \ne p. \text { Thus, } n = 1 \text { and } y \ne 0.[/MATH]
So you can put yes in the cell for u = 1. Put no in all the other 1 cells, and put no in y’s zero cell.
Now there are two things that look a bit interesting, but I I do not know whether they lead anywhere.
One is that [MATH]t(100w + 10h + y) = 1100b + 10n + p.[/MATH]
The other is that [MATH]carry-digit + b + u = n \le 9.[/MATH] Why?
n should be u i think.n=1 and y≠0.If y=0⟹t∗y=0+10∗0=y≠p. Thus, n=1 and y≠0.\displaystyle \text {If } y = 0 \implies t * y = 0 + 10 * 0 = y \ne p. \text { Thus, } n = 1 \text { and } y \ne 0.
So you can put yes in the cell for u = 1.
if u=1 then u*h=p+carry*10 (this cannot happen ).
why? did not understandcarry−digit+b+u=n≤9.carry−digit+b+u=n≤9.\displaystyle carry-digit + b + u = n \le 9. Why?
You are correct. I made an error. Good work catching it. I apologize; that was sloppy of me.n should be u i think.
if u=1 then u*h=p+carry*10 (this cannot happen ).
WHY
NUT
T times WHY = BBNP
U times WHY = BYPY
N times WHY = BUHA
BBNP + 10 * BYPY + 100 * BUHA = BNEPBP
[MATH]u * y = y + 10\alpha \implies \\ y = 0 \text { and } u \ne 0;\\ y \ne 1 \text { and } u = 1;\\ y = 2 \text { and } u = 6;\\ y = 4 \text { and } u = 6;\text { or }\\ y = 8 \text { and } u = 6.[/MATH][MATH]\text {If } y = 0 \implies p = ty = 0 \implies p = y \text { impossible} \implies y \ne 0.[/MATH]
Put no in the y = 0 cell.
[MATH]\text {If } u = 1 \implies u * y = y \text { and } p = 1 * h = h \implies p = h \text { impossible} \implies u \ne 1.[/MATH]
[MATH]\therefore u = 6 \text { and } \\ y = 2 \text { and } \alpha = 1;\\ y = 4 \text { and } \alpha = 2; \text { or}\\ y = 8 \text { and } \alpha = 4.[/MATH]Put yes in the u = 6 cell and no in all the other u cells. Put no in all the y cells except 2, 4, and 8.
There is a carry digit from the thousands column. Call it [MATH]\gamma^*.[/MATH]why? did not understand
There is a carry digit from the ten thousands column. Call it [MATH]\delta^*.[/MATH]
[MATH]\gamma^* + b + u = n + 10\delta^* \text { and } \delta^* + b = b.[/MATH]
Follow that? And obviously
[MATH]\delta^* + b = b \implies \delta^* = 0 \implies 9 \ge n = \gamma^* + b + u = \gamma^* + b + 6 \implies 1 \le \gamma^* + b \le 3.[/MATH]
You can put no in the b cells for all values > 3. You can put no in the n cells for all values < 7.
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absolutely i thought this way.If u=1⟹u∗y=y and p=1∗h=h⟹p=h impossible⟹u≠1.If u=1⟹u∗y=y and p=1∗h=h⟹p=h impossible⟹u≠1.\displaystyle \text {If } u = 1 \implies u * y = y \text { and } p = 1 * h = h \implies p = h \text { impossible} \implies u \ne 1.
∴u=6 and y=2 and α=1;y=4 and α=2; ory=8 and α=4.∴u=6 and y=2 and α=1;y=4 and α=2; ory=8 and α=4.\displaystyle \therefore u = 6 \text { and } \\ y = 2 \text { and } \alpha = 1;\\ y = 4 \text { and } \alpha = 2; \text { or}\\ y = 8 \text { and } \alpha = 4.
Put yes in the u = 6 cell and no in all the other u cells. Put no in all the y cells except 2, 4, and 8.
But if U={ 3,7,9} and Y=5 . this can also happen
No. I must admit I saw the negative, but did not explain it. That is the problem with trying to explain these puzzles. You “see” something as obvious and then forget to explain it because you imagine it is obvious.absolutely i thought this way.
But if U={ 3,7,9} and Y=5 . this can also happen
The units digits of ty, uy, and ny are p, y, and a. But if y = 5, this is impossible because p, y, and a are different numbers.
any other lead U got .No. I must admit I saw the negative, but did not explain it. That is the problem with trying to explain these puzzles. You “see” something as obvious and then forget to explain it because you imagine it is obvious.
The units digits of ty, uy, and ny are p, y, and a. But if y = 5, this is impossible because p, y, and a are different numbers.
You really have a lot of information now
u = 6
y = 2, 4, or 8
b = 1, 2, or 3
n = 7, 8, or 9
So experiment.
[MATH]\text {If } n = 9 \text { and } y = 2 \implies n + y = 11 = 1 + 10 \implies b = 1.[/MATH]
[MATH]\text {If } n = 9 \text { and } y = 4 \implies n + y = 13 = 3 + 10 \implies b = 3.[/MATH]
[MATH]\text {If } n = 9 \text { and } y = 4 \implies 9 * 4 = 36 = 6 + 30 \implies a = u \text {: IMPOSSIBLE.}[/MATH]
[MATH]\text {If } n = 9 \text { and } y = 8 \implies n + y = 17 = 7 + 10 \implies b = 7 \text {: IMPOSSIBLE.}[/MATH]
[MATH]\text {If } n = 9, \ y = 2, \text { and } b = 1 \implies 2t \ne t \implies t \ne 0.[/MATH]
[MATH]\therefore w > 2 \text { and } t > 2 \implies 100w * t \ge 1200 > 1199 \implies b \ne 1: \text { IMPOSSIBLE.}[/MATH]
[MATH]\therefore n \ne 9.[/MATH]
u = 6
y = 2, 4, or 8
b = 1, 2, or 3
n = 7, 8, or 9
So experiment.
[MATH]\text {If } n = 9 \text { and } y = 2 \implies n + y = 11 = 1 + 10 \implies b = 1.[/MATH]
[MATH]\text {If } n = 9 \text { and } y = 4 \implies n + y = 13 = 3 + 10 \implies b = 3.[/MATH]
[MATH]\text {If } n = 9 \text { and } y = 4 \implies 9 * 4 = 36 = 6 + 30 \implies a = u \text {: IMPOSSIBLE.}[/MATH]
[MATH]\text {If } n = 9 \text { and } y = 8 \implies n + y = 17 = 7 + 10 \implies b = 7 \text {: IMPOSSIBLE.}[/MATH]
[MATH]\text {If } n = 9, \ y = 2, \text { and } b = 1 \implies 2t \ne t \implies t \ne 0.[/MATH]
[MATH]\therefore w > 2 \text { and } t > 2 \implies 100w * t \ge 1200 > 1199 \implies b \ne 1: \text { IMPOSSIBLE.}[/MATH]
[MATH]\therefore n \ne 9.[/MATH]
yes t not equal to zero.2t≠t⟹t≠0.If n=9, y=2, and b=1⟹2t≠t⟹t≠0.\displaystyle \text {If } n = 9, \ y = 2, \text { and } b = 1 \implies 2t \ne t \implies t \ne 0.
so, w*t+ carry =b+b*10 and bb can be 11 or 22 or 33 or ...100w∗t≥1200>1199⟹b≠1: IMPOSSIBLE.∴w>2 and t>2⟹100w∗t≥1200>1199⟹b≠1: IMPOSSIBLE.\displaystyle \therefore w > 2 \text { and } t > 2 \implies 100w * t \ge 1200 > 1199 \implies b \ne 1: \text { IMPOSSIBLE.}
I understood that place value of w is 100 so 100 *w * t gives a value .Now least value of w=3 and t=4 so 1200
But what does 1199 mean. Yes it is one less than 1200. but why did u bring it here and How it is implying b not equal to 1 ?
My thought process of b not equal to 1 is Y=2 , we can get 11 by w*t+ carry--> 5*2 + carry (1) but 2 is taken and 3*2+5 also 2 is taken .SO b is not equal to 1.
For other combinations of U=6, (N,Y)= (9,2) or (9,4) or (9,8) or (8,4) or (8,2) or (7,8) these are not valid
I am trying with N,Y=(7,4) or (7,2).
First, we could have found t is neither zero nor one way back at the start, going to Subhotosh’s point that there were things easy to see from the beginning. Better late than never.yes t not equal to zero.
so, w*t+ carry =b+b*10 and bb can be 11 or 22 or 33 or ...
I understood that place value of w is 100 so 100 *w * t gives a value .Now least value of w=3 and t=4 so 1200
But what does 1199 mean. Yes it is one less than 1200. but why did u bring it here and How it is implying b not equal to 1 ?
My thought process of b not equal to 1 is Y=2 , we can get 11 by w*t+ carry--> 5*2 + carry (1) but 2 is taken and 3*2+5 also 2 is taken .SO b is not equal to 1.
For other combinations of U=6, (N,Y)= (9,2) or (9,4) or (9,8) or (8,4) or (8,2) or (7,8) these are not valid
I am trying with N,Y=(7,4) or (7,2).
Second, it is not true that we can conclude, from b = 1, y = 2, w > 2, t > 2, and w not equal to t, that the least value of w is 3 and the least value of t is 4. It is possible that t may be 3 and the least value of w is 4.
[MATH]t > w \ge 3 \implies t \ge 4 \text { and } tw \ge 12. \text { And } w > t \ge 3 \implies w \ge 4 \text { and } tw \ge 12.[/MATH]
So, whatever the relationship between t and w, [MATH]t * 100w \ge 1200.[/MATH]
Third, I want to compare that product to BBNP given that b = 1. Ignoring what n and p might be for the moment, that number must not exceed 1199. Now it is true that we are assuming n = 9 so p cannot exceed 8. Being precise, I can conclude that BBNP cannot exceed 1198. Bit I do not need to be precise. The number cannot exceed 1199 if b = 1, but must exceed 1199 if b = 1. Just a different way of getting there.
Fourth, I cannot tell whether your exclusion of n = 8 is correct BECAUSE YOU GAVE NO REASONING.
Fifth, if n = 7, then b = 1.
niceThird, I want to compare that product to BBNP given that b = 1. Ignoring what n and p might be for the moment, that number must not exceed 1199. Now it is true that we are assuming n = 9 so p cannot exceed 8. Being precise, I can conclude that BBNP cannot exceed 1198. Bit I do not need to be precise. The number cannot exceed 1199 if b = 1, but must exceed 1199 if b = 1. Just a different way of getting there.
I cannot tell whether your exclusion of n = 8 is correct BECAUSE YOU GAVE NO REASONING.
n=8,y=4
n+y=12 =>b=2
n*y=8*4=32->A=2 but b=2
n=8,y=2
n+y=10 -> b=0
n*y=16 => A=6 but U=6
Fifth, if n = 7, then b = 1.
okay now in N,Y=(7,4)
B=1
N*y=7*4=28=>A=8
BBNP=1179
w*T*100 = 2*3*100=600 (assume w=2, T=3)
So, 600 <_ 1179 .
SO this N,Y=(7,4) combo is valid ? right and then we need to further check other things.
this combo (7,2) is still left to be checked.
Here is how I confirmed your conclusion that n = 7. We rejected n = 9. Test n = 8.
[MATH]t * (100w + 10h + y) = 1000b + 100b + 10n + p\\ 10u* (100w + 10h + y) = 10000b + 1000y + 100p + 10y + 0 \\ 100n(100w + 10h + y) = 100000b + 10000u + 1000h + 100a + 0 + 0 \text { and}\\ 100000b + 10000(b + u) + 1000(b + y + h) + 100(b + p + a) + 10(n + y) + p =\\ 100000b + 10000n + 1000e + 100p + 10b + p.[/MATH]And we know from the very start that w > 0, n > 0, b > 0, and t > 1.
We figured out that u = 6, n = 7, 8, or 9, b = 1, 2, or 3, and r = 2, 4, or 8. (Assuming no error.)
We then deduced n = 7 or 8, meaning b = 1 or 2. (Again assuming no error).
Let’s try n = 8 and b = 1.
[MATH]\therefore y > 1 \implies n + y > 8 + 1 \implies 8 + y = b + 10 = 1 + 10 \implies y = 3: \text { IMPOSSIBLE.}[/MATH]
Try n = 8 and b = 2, which means y = 4.
[MATH]8 + 4 = 12 = 2 + 10 = b + 10.[/MATH] OK
[MATH]p + 10? = t * y = 4t.\\ \text {If } t = 3 \implies p = 2 = b \implies t = 5, \ 7, \text { or } 9.[/MATH]
[MATH]t = 7 \implies p = 8 = n \implies t \ne 7.[/MATH]
[MATH]t = 9 \implies p = 6 = u \implies t \ne 9.[/MATH]
[MATH]t = 5 \implies p = 0 \text { and } 2 + 5h = 8 + 10? \implies 5h = 6 + 10?: \text { IMPOSSIBLE.}[/MATH]
So n = 7 and b = 1.
Thus y = 4.
[MATH]170000 < 100000b + 10000n + 1000e + 100p + 10b + p < 179999\\ 760 < 100n + 10u + t < 770\\ \therefore 170000 \div 770 < (100000b + 10000n + 1000e + 100p + 10b + p) \div (100n + 10u + t) < 179999 \div 760 \implies\\ 220 < 100w + 10h + 4 < 237 \implies w = 2 \text { and } h = 3.[/MATH]So 100w + 10h + y = 234.
Now we can use brute force. We also now know t = 5, 8, or 9. Try them.
234 * 765 = 1170 + 14040 + 163800 = 179010. That gives us
p = 0,
b = 1,
w = 2,
y = 4,
t = 5,
u = 6,
n = 7,
a = 8,
e = 9
[MATH]t * (100w + 10h + y) = 1000b + 100b + 10n + p\\ 10u* (100w + 10h + y) = 10000b + 1000y + 100p + 10y + 0 \\ 100n(100w + 10h + y) = 100000b + 10000u + 1000h + 100a + 0 + 0 \text { and}\\ 100000b + 10000(b + u) + 1000(b + y + h) + 100(b + p + a) + 10(n + y) + p =\\ 100000b + 10000n + 1000e + 100p + 10b + p.[/MATH]And we know from the very start that w > 0, n > 0, b > 0, and t > 1.
We figured out that u = 6, n = 7, 8, or 9, b = 1, 2, or 3, and r = 2, 4, or 8. (Assuming no error.)
We then deduced n = 7 or 8, meaning b = 1 or 2. (Again assuming no error).
Let’s try n = 8 and b = 1.
[MATH]\therefore y > 1 \implies n + y > 8 + 1 \implies 8 + y = b + 10 = 1 + 10 \implies y = 3: \text { IMPOSSIBLE.}[/MATH]
Try n = 8 and b = 2, which means y = 4.
[MATH]8 + 4 = 12 = 2 + 10 = b + 10.[/MATH] OK
[MATH]p + 10? = t * y = 4t.\\ \text {If } t = 3 \implies p = 2 = b \implies t = 5, \ 7, \text { or } 9.[/MATH]
[MATH]t = 7 \implies p = 8 = n \implies t \ne 7.[/MATH]
[MATH]t = 9 \implies p = 6 = u \implies t \ne 9.[/MATH]
[MATH]t = 5 \implies p = 0 \text { and } 2 + 5h = 8 + 10? \implies 5h = 6 + 10?: \text { IMPOSSIBLE.}[/MATH]
So n = 7 and b = 1.
Thus y = 4.
[MATH]170000 < 100000b + 10000n + 1000e + 100p + 10b + p < 179999\\ 760 < 100n + 10u + t < 770\\ \therefore 170000 \div 770 < (100000b + 10000n + 1000e + 100p + 10b + p) \div (100n + 10u + t) < 179999 \div 760 \implies\\ 220 < 100w + 10h + 4 < 237 \implies w = 2 \text { and } h = 3.[/MATH]So 100w + 10h + y = 234.
Now we can use brute force. We also now know t = 5, 8, or 9. Try them.
234 * 765 = 1170 + 14040 + 163800 = 179010. That gives us
p = 0,
b = 1,
w = 2,
y = 4,
t = 5,
u = 6,
n = 7,
a = 8,
e = 9