WATCH 
1 page 1100 views  Post Reply 
RunRyder  RunRyder 
red_z06
rrProfessor Dumont, NJ 
Let see how smart some of you guys are. Find this 3 digit number that satisfies following condition.1. 10th place number is 62. Switching 100th and 1th number causes the new number to be smaller than the original by 297.3. Switching 100th and 1th of the original number and adding 179 to it will turn the new number to have same 100th 10th and 1th digits. For example 111, 222 etc. 
SHARE PM EMAIL HOMEPAGE GALLERY Attn:RR Quote 
utahbob
rrVeteran St. George Utah 💎Sustaining Member 
huh? aahaahaah90210? 
SHARE PM EMAIL GALLERY Attn:RR Quote 
baldgeek
rrApprentice New Zealand 

SHARE PM EMAIL Attn:RR Quote 
PsychoZ
rrApprentice Northern, CA 

SHARE PM EMAIL GALLERY Attn:RR Quote 
JoseA
rrNovice South Chicago 

SHARE PM EMAIL GALLERY Attn:RR Quote 
red_z06
rrProfessor Dumont, NJ 

SHARE PM EMAIL HOMEPAGE GALLERY Attn:RR Quote 
PsychoZ
rrApprentice Northern, CA 
8765309 
SHARE PM EMAIL GALLERY Attn:RR Quote 
titillater
rrApprentice Duncan B.C 

SHARE PM EMAIL GALLERY Attn:RR Quote 
Climax
rrVeteran West London, United Kingdom 
100Y + 60 + X +297 = 100X + 60 + YWhich simplifies to,X  Y = 3 (Remember this as equation 1)OK, you just need one more equation relating X to Y and you can then solve them as a simultaneous pairFrom point (3)100Y + 60 + X + 179 = 100Z +10Z + ZWhere Z represents the digits in either 111 to 999 (i.e. ZZZ, as described in point 3 of the problem)This simplifies to,100Y + 60 + X + 179 = 111Z (Remember this as equation 2)You still need to figure out Z, here's what I did...We know that in the answer the middle digit is "6" (as we have been told this in the question)We also know (in symbolic terms) that Y6X + 179 = ZZZNow consider the middle digits, here 6 + 7 = 13. So it would appear that the middle Z might be a "3" with a carry that ripples up (13 = 10 + 3, or 3 carry 1). It also possible that there was a "carry up" from the lower down X + 9 component, in which case the middle Z would be a 4.So we now know that Z is either 3 or 4. Now if you think about it, it can't be a "3" as there is always a carry from the lowest digit. Why, well if Z is 3 or 4, then the X + 9 component will always generate a carry!So Z = 4This means that equation 2 becomes,100Y + 60 + X + 179 = 444Which simplifies to,100Y + X = 205 (Remember this as equation 3)We now have 2 equations in X and Y100Y + X = 205 X  Y = 3 (this was equation 1 from above)Let's find X...X = 3 + YThen substitute in to equation 3,100Y + (3 + Y) = 205 101Y = 202 Y = 2Then as X = 3 + Y,X = 5Which means that the answer is 562I'm sure there must be a simpler way, any ideas? 
SHARE PM EMAIL HOMEPAGE GALLERY Attn:RR Quote 
Shawn Pierce
rrVeteran Galliano, Louisiana 

SHARE PM EMAIL GALLERY Attn:RR Quote 
Climax
rrVeteran West London, United Kingdom 

SHARE PM EMAIL HOMEPAGE GALLERY Attn:RR Quote 
mr dan
rrVeteran Stockton Calif 
But can you you fly inverted while holding your TX behind your back, hanging upside down and drinking a soda? 
SHARE PM EMAIL GALLERY Attn:RR Quote 
red_z06
rrProfessor Dumont, NJ 

SHARE PM EMAIL HOMEPAGE GALLERY Attn:RR Quote 
red_z06
rrProfessor Dumont, NJ 
B = 1th digitB6A+179=CCCSince A is 100th digit, it can't be zero. If A can't be zero, then C=4 (for the values of A 1 through 9)So, 444179=265B=2 A=5Answer is 562. 
SHARE PM EMAIL HOMEPAGE GALLERY Attn:RR Quote 
WATCH 
1 page 1100 views  Post Reply 
RunRyder  RunRyder 
Thursday, October 22  7:49 am  Copyright © 20002020 RunRyder EMAIL • Enable Cookies