@bodhi_vriksha said:
Yes, i do remember this one:) This problem appeared in RMO 2006.
Team BV - Vineet
attached fig :
@jain4444 said:Find the sum of all possible values of A + C + F + J if each letter denotes a single distinct digit and A + BC + DEF = GHIJ.via - kamal/vineet sir
1, 0, 9 are already fixed for G, H and D.
And as sum of all the digits is 45 which needs to be divided among two parts (on both sides of =), but they can't be equal as 45 is odd.
So clearly there difference (of the form 9k) has to be odd. That is, the carry 'k' has to be odd i.e. 1 or 3.
Carry of '1' means, sum of digits of LHS = 27 and that of RHS = 18.
Here in RHS we have already fixed two letters as 1 and 0, so sum of remaining two must be 17 which can be achieved only by 9 + 8.
But '9' has already been used, so this case is not possible.
Carry of '3' means, sum of digits of LHS = 36 and that of RHS = 9.
Here in RHS we have already fixed two letters as 1 and 0, so sum of remaining two must be 8 which can be achieved by 2 + 6 or 3 + 5 alternately.
And in each addition step, we need '1' carry as hundred's place 9 is getting '1' carry. And by adding ten's place digits, we need exactly '1' carry to be added in 9. So, certainly, by adding unit digits we must get a carry of '1' so that total requirement of carry of '3' is met.
Thus with all four possible values (2, 3, 5, 6) of unit digit of sum i.e. J, we can have
(i) A + C + F + J = 12 + 2 = 14 (12 can be achieved by 3 + 4 + 5)
(ii) A + C + F + J = 13 + 3 = 16 (13 can be achieved by 2 + 4 + 7 or 2 + 5 + 6)
(iii) A + C + F + J = 15 + 5 = 20 (15 can be achieved by 2 + 6 + 7 or 3 + 4 + 8)
(iv) A + C + F + J = 16 + 6 = 22 (16 can be achieved by 3 + 5 + 8 or 4 + 5 + 7)
So basically there are many cases but all possible values of A + C + F + J are 14, 16, 20 and 22 only. :)
Team BV - Kamal Lohia
And as sum of all the digits is 45 which needs to be divided among two parts (on both sides of =), but they can't be equal as 45 is odd.
So clearly there difference (of the form 9k) has to be odd. That is, the carry 'k' has to be odd i.e. 1 or 3.
Carry of '1' means, sum of digits of LHS = 27 and that of RHS = 18.
Here in RHS we have already fixed two letters as 1 and 0, so sum of remaining two must be 17 which can be achieved only by 9 + 8.
But '9' has already been used, so this case is not possible.
Carry of '3' means, sum of digits of LHS = 36 and that of RHS = 9.
Here in RHS we have already fixed two letters as 1 and 0, so sum of remaining two must be 8 which can be achieved by 2 + 6 or 3 + 5 alternately.
And in each addition step, we need '1' carry as hundred's place 9 is getting '1' carry. And by adding ten's place digits, we need exactly '1' carry to be added in 9. So, certainly, by adding unit digits we must get a carry of '1' so that total requirement of carry of '3' is met.
Thus with all four possible values (2, 3, 5, 6) of unit digit of sum i.e. J, we can have
(i) A + C + F + J = 12 + 2 = 14 (12 can be achieved by 3 + 4 + 5)
(ii) A + C + F + J = 13 + 3 = 16 (13 can be achieved by 2 + 4 + 7 or 2 + 5 + 6)
(iii) A + C + F + J = 15 + 5 = 20 (15 can be achieved by 2 + 6 + 7 or 3 + 4 + 8)
(iv) A + C + F + J = 16 + 6 = 22 (16 can be achieved by 3 + 5 + 8 or 4 + 5 + 7)
So basically there are many cases but all possible values of A + C + F + J are 14, 16, 20 and 22 only. :)
Team BV - Kamal Lohia
@bodhi_vriksha said:1, 0, 9 are already fixed for G, H and D.And as sum of all the digits is 45 which needs to be divided among two parts (on both sides of =), but they can't be equal as 45 is odd.So clearly there difference (of the form 9k) has to be odd. That is, the carry 'k' has to be odd i.e. 1 or 3.Carry of '1' means, sum of digits of LHS = 27 and that of RHS = 18.Here in RHS we have already fixed two letters as 1 and 0, so sum of remaining two must be 17 which can be achieved only by 9 + 8. But '9' has already been used, so this case is not possible.Carry of '3' means, sum of digits of LHS = 36 and that of RHS = 9.Here in RHS we have already fixed two letters as 1 and 0, so sum of remaining two must be 8 which can be achieved by 2 + 6 or 3 + 5 alternately.And in each addition step, we need '1' carry as hundred's place 9 is getting '1' carry. And by adding ten's place digits, we need exactly '1' carry to be added in 9. So, certainly, by adding unit digits we must get a carry of '1' so that total requirement of carry of '3' is met.Thus with all four possible values (2, 3, 5, 6) of unit digit of sum i.e. J, we can have(i) A + C + F + J = 12 + 2 = 14 (12 can be achieved by 2 + 3 + 7 or 2 + 4 + 6 or 3 + 4 + 5)(ii) A + C + F + J = 13 + 3 = 16 (13 can be achieved by 2 + 3 + 8 or 2 + 4 + 7 or 2 + 5 + 6 or 3 + 4 + 6)(iii) A + C + F + J = 15 + 5 = 20 (15 can be achieved by 2 + 5 + 8 or 2 + 6 + 7 or 3 + 4 + 8 or 3 + 5 + 7 or 4 + 5 + 6)(iv) A + C + F + J = 16 + 6 = 22 (16 can be achieved by 2 + 6 + 8 or 3 + 5 + 8 or 3 + 6 + 7 or 4 + 5 + 7)So basically there are many cases but all possible values of A + C + F + J are 14, 16, 20 and 22 only. Team BV - Kamal Lohia
Sir, don't the digits repeat here?? Last digit in 8+3+2 = (3) --> Repeating, whereas we are required to find distinct digits! (As per question)
@jain4444 said:Find the sum of all possible values of A + C + F + J if each letter denotes a single distinct digit and A + BC + DEF = GHIJ.via - kamal/vineet sir
i am getting only one possible value = 16
it is clear that g= 1, d= 9 , h = 0...after that did trial and error mostly, getting a,c,f = 2,4,7 in some order and j = 3
pls share, if you have a shorter approach
@RDN said:i am getting only one possible value = 16it is clear that g= 1, d= 9 , h = 0...after that did trial and error mostly, getting a,c,f = 2,4,7 in some order and j = 3pls share, if you have a shorter approach
Yeah, same method! Had to see through all combinations and left after that.
@RDN said:@ScareCrow28 ek aur hai...2 + 7 + 6 +5 = 20
Good! 😃 The thing is we are checking one by one which is obviously not a right way to solve this problem or else the question is a waste :P
@amresh_maverick said:attached fig :
Remember that sum of all the exterior angles of a planar polygon is 360 degrees.
Here each convex corner (with internal angle 90 degrees) has exterior angle of 90 degrees.
And each concave corner (with internal angle of 270 degree) has exterior angle of -90 degrees.
So using the given data, we can write
90(25) - 90(n) = 360
i.e. 25 - n = 4 and n = 21 (C)
Team BV - Kamal Lohia
@ScareCrow28 that's true, but the best we can do is lend some order to the data, to make our lives easier...oc, the wisest option would be to leave the question
once we have removed 0,1,9 for the last column summation we have the following selection to make- select 3 out of (2/8), (3/7) , (4/6) , 5 ... from here on it is hell, i agree
however, there is usually a smarter approach to alphametics, quant thread guru logon ko bulao...
@ScareCrow28 said:Sir, don't the digits repeat here?? Last digit in 8+3+2 = (3) --> Repeating, whereas we are required to find distinct digits! (As per question)
Oh yes! You are right. I overlooked that.
Revised cases will be:
12 = 3 + 4 + 5
13 = 2 + 4 + 7 = 2 + 5 + 6
15 = 2 + 6 + 7 = 3 + 4 + 8
16 = 3 + 5 + 8 = 4 + 5 + 7
Thanks again for correction. :)
my solution above
Team BV - Kamal Lohia
Revised cases will be:
12 = 3 + 4 + 5
13 = 2 + 4 + 7 = 2 + 5 + 6
15 = 2 + 6 + 7 = 3 + 4 + 8
16 = 3 + 5 + 8 = 4 + 5 + 7
Thanks again for correction. :)
Team BV - Kamal Lohia
@bodhi_vriksha said:Oh yes! You are right. I overlooked that.Revised cases will be: 12 = 3 + 4 + 513 = 2 + 4 + 7 = 2 + 5 + 615 = 2 + 6 + 7 = 3 + 4 + 816 = 3 + 5 + 8 = 4 + 5 + 7Thanks again for correction. Team BV - Kamal Lohia
sir, 20 is also one of the sums. (2 + 7 + 6) = 5 ( Last digit)
@jain4444 said:Find the sum of all possible values of A + C + F + J if each letter denotes a single distinct digit and A + BC + DEF = GHIJ.via - kamal/vineet sir
My go at this one...
If the remainder of LHS and RHS is r with 9, then r must be 0 because 2r =45mod9=0mod9.
Also since we must have atleast two carries while operating the LHS items, the RHS digits must add to 9 and LHS digits must add to 36. Hence total carries = 3.
We also know that D=9, G=1, and H=0. So,I+J=8=> (I,J)=(2,6), (6,2), (3,5), (5,3)
So, the only possible values of RHS are 1026, 1062, 1035, and 1053.
So, the only possible values of RHS are 1026, 1062, 1035, and 1053.
For j=6, (a,c,f)= Permutation (3,5,8) => Sum =22
For j=2, (a,c,f)= Permutation (3,5,4) => Sum =14
For j=5, (a,c,f)= Permutation (2,6,7) => Sum =20
For j=3, (a,c,f)= Permutation (2,4,7) => Sum =16
So, total sum = 14+16+20+22=72
Team BV - Vineet
@ScareCrow28 said:sir, 20 is also one of the sums. (2 + 7 + 6) = 5 ( Last digit)
I have already taken the sum 20. And 2 + 6 + 7 is same as 2 + 7 + 6. Isn't it? :)
We are just looking at the sum right now, not the arangements. Otherwise 6 + 7 + 2 will also make a different case. :)
Team BV - Kamal Lohia
We are just looking at the sum right now, not the arangements. Otherwise 6 + 7 + 2 will also make a different case. :)
Team BV - Kamal Lohia
@bodhi_vriksha said:I have already taken the sum 20. And 2 + 6 + 7 is same as 2 + 7 + 6. Isn't it? We are just looking at the sum right now, not the arangements. Otherwise 6 + 7 + 2 will also make a different case. Team BV - Kamal Lohia
Yeah, overlooked that before, sir! 😃
Determine all positive integers n such that exactly n/3 positive integers are ( No OA)
@ScareCrow28 said:Determine all positive integers n such that exactly n/3 positive integers are
n * 1/2 * 2/3 = n/3
so , "n" should have only powers of 2 and 3
n = 2^x * 3^y
now "x" and "y" must have some range
@ScareCrow28 bhai kuch gadbad kardi kya ?
@jain4444 said:n * 1/2 * 2/3 = n/3 so , "n" should have only powers of 2 and 3 n = 2^x * 3^y now "x" and "y" must be some range @ScareCrow28 bhai kuch gadbad kardi kya ?
Sir aap aur galat? 
Looks absolutely flawless to me! Should be the right answer 
Looks absolutely flawless to me! Should be the right answer @gautam22 said:if 1 is not relative prime to any number den i guess ans shud be only number 3.......batao kuch theek lag raha hai to proof likhoon
1 is relative prime to any other positive integer
If GCD(a, b) = 1, where a and b are positive integers, then a and b are co-prime to each other or relative prime numbers
@gautam22 said:if 1 is not relative prime to any number den i guess ans shud be only number 3.......batao kuch theek lag raha hai to proof likhoon
Bhai, we will take 1 as relative prime to a number. @jain4444 sir ne upar sahi kia hai.. :)