Strange ...... just a fleeting look tells meView attachment 27875
I tried but nothing came .there is nothing to Show
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.
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 ).
There is a carry digit from the thousands column. Call it [MATH]\gamma^*.[/MATH]why? did not understand
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.
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
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.
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.}
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).
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.
Fifth, if n = 7, then b = 1.