[ashish01]

Another great thread which is gonna be reall helpful for those who were snooring till now.
This looks like an LR puzzle but surely will land-up on QA section.


A+B+C+D=D+E+F+G=G+H+I=17 where each letter represent a number from 1 to 9. Find out what does letter D and G represent if letter A= 4.

Is there a definite approach to it rather than trying all the combinations?

[gsnindia]

Originally Posted by ashish01

A+B+C+D=D+E+F+G=G+H+I=17 where each letter represent a number from 1 to 9. Find out what does letter D and G represent if letter A= 4.

Did not find any Good Arroach But have some good calculation

it is like that

a+b+c+d=17
d+e+f+g=17
g+h+i=17

Start with G+h+i taking any number say 9+7+1 we will decide G later

G can take either 9,7,1 we make D+e+f can take 8,10,16, we cannot take 9,7 as our sum will require shorter no only. so we take 1 as G
so D+E+F = 16 , 8+5+3 But we have not decided what D can take so we first add A+B+C we are left out with number 6 and 2 as A=4 known so 6+4+2 = 12 D can take only 17-12 =5

Hence D,G can take 5,1

another easy approach (Striked immediatly after i have solved the problem.) is

we have 1+2+3+4+5+6+7+8+9 = 45 any 2 number are repeated
a+b+c+d=17
d+e+f+g=17
g+h+i=17

which make a total of 51 so the addional no require is (17*3)-45 = 6

combination can be only (1,5) , (2,4) sicne A=4 so we left with only choice (5,1) -- (1,5) Not possible try . No need to solved as i did. but anyway keeping as what approach i used to help other understand.

[kabir1224]

Originally Posted by gsnindia

Did not find any Good Arroach But have some good calculation

it is like that

a+b+c+d=17
d+e+f+g=17
g+h+i=17

Start with G+h+i taking any number say 9+7+1 we will decide G later

G can take either 9,7,1 we make D+e+f can take 8,10,16, we cannot take 9,7 as our sum will require shorter no only. so we take 1 as G
so D+E+F = 16 , 8+5+3 But we have not decided what D can take so we first add A+B+C we are left out with number 6 and 2 as A=4 known so 6+4+2 = 12 D can take only 17-12 =5

Hence D,G can take 5,1

another easy approach (Striked immediatly after i have solved the problem.) is

we have 1+2+3+4+5+6+7+8+9 = 45 any 2 number are repeated
a+b+c+d=17
d+e+f+g=17
g+h+i=17

which make a total of 51 so the addional no require is (17*3)-45 = 6

combination can be only (1,5) , (2,4) sicne A=4 so we left with only choice (5,1) -- (1,5) Not possible try . No need to solved as i did. but anyway keeping as what approach i used to help other understand.

Great approach man really. Keep up the good work. One ques from my side
if x+y+z=10 then how many solutions does this equation has if x,yn z are real nos.?

[bharat singh]

Originally Posted by gsnindia

The Answer to This Question is As below

we first make them to the power of 2

2^4=16 , hence 2^20 + 2^15 = 2^15(2^5 +1) = 2^15(31) hence it is divisible by 31

is tht 33.
i mean i got it 33 yaar, hey bro u made a mistake in addition ....

anyways u guys r doing a gr88 job by introducing this thread,i mean its a kind of
gr88 help to ppl like me who r pathetic in quant.i too shall contribute to make it successful. cheers.......

[warrior]

Originally Posted by ashish01

Another great thread which is gonna be reall helpful for those who were snooring till now.
This looks like an LR puzzle but surely will land-up on QA section.


A+B+C+D=D+E+F+G=G+H+I=17 where each letter represent a number from 1 to 9. Find out what does letter D and G represent if letter A= 4.

Is there a definite approach to it rather than trying all the combinations?

add alll we get 45+ D+G =51

d+G= 51 = 1,5 or 2,4 or 3,3

3,3 and 2,4 not possible

so 1,5

E+F= 11

also if G= 1 H+I= 16

now if G= 5 H,I =9,3

E,F can be 6,5 7,4 8,3 9,2 but none is possible so G=1 D=5

[gsnindia]

Originally Posted by kabir1224

if x+y+z=10 then how many solutions does this equation has if x,yn z are real nos.?

Can Any one put some easy light for us to see how we can get the solution for such questions

[nairpraveenk]

Awesome thread!! Just what we need to brush up our basics!!

[mr.s.k.abhi]

Originally Posted by kabir1224

if x+y+z=10 then how many solutions does this equation has if x,yn z are real nos.?

Well my QA is worse but i think I can give this one a try... PLz correct me if i m wroung or some one finds a easy way out....

WELL x + y + z = 10

My approch is based on assumption and susbtitution..
asuming X=0
y+z=10
for this y & z can have 10 st of values from (0,10) to (10,0)
similarly assuming X=1
y+z=9 ===> y,z have 9 pair of values

Going by this approach
total no. of values for (x,y,z) = 10+9+8+7+6+5+4+3+2+1=55

So i think 55 is the answer.

__________________________________________________ _____________
NEVER SAY DIE...........:smilecol: :snipersm:

[maverickisneo]

Originally Posted by mr.s.k.abhi

Well my QA is worse but i think I can give this one a try... PLz correct me if i m wroung or some one finds a easy way out....

WELL x + y + z = 10

My approch is based on assumption and susbtitution..
asuming X=0
y+z=10
for this y & z can have 10 st of values from (0,10) to (10,0)
similarly assuming X=1
y+z=9 ===> y,z have 9 pair of values

Going by this approach
total no. of values for (x,y,z) = 10+9+8+7+6+5+4+3+2+1=55

So i think 55 is the answer.

if x,y and z are real numbers then there could be infinite solutions.However, if x,y and z are positive integers then i am getting 66 as answer.plz correct me if i am wrong.


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[warrior]

Originally Posted by maverickisneo

if x,y and z are real numbers then there could be infinite solutions.However, if x,y and z are positive integers then i am getting 66 as answer.plz correct me if i am wrong.


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If x,y,z are real - infinite

If x,y,z non negative - 66

if x,y,z positive then - 36