[nishant.pilania]

Originally Posted by apocalyptic

Find the value of(7/9)*(26/27)*(63/65)*(124/126).........10 terms ?

whats the series
i'm not getting the relation
plz help me by explaining ur series a bit

[target_iima]

The product of two numbers ‘231’ and ‘ABA’ is ‘BA4AA’ in a certain base system (where base is less than 10), where A and B are distinct digits. What is the base of that system?pls suggest the full solution...

[shreyans_nitt]

Originally Posted by target_iima

The product of two numbers ‘231’ and ‘ABA’ is ‘BA4AA’ in a certain base system (where base is less than 10), where A and B are distinct digits. What is the base of that system?pls suggest the full solution...

let the base be k(1 + 3k + 2k^2)*(A + Bk + Ak^2) = (A + Ak + 4k^2 + Ak^3 + Bk^4)(B+2A)k + (3B +3A -4)k^2 + (2A+2B)k^3 + (2A-B)k^4 = 0what to do after this??

[pquanta]

Originally Posted by target_iima

The product of two numbers ‘231’ and ‘ABA’ is ‘BA4AA’ in a certain base system (where base is less than 10), where A and B are distinct digits. What is the base of that system?pls suggest the full solution...

base 6.
A =1, B = 4.
2*36 + 3*6 +1 = 91
91(Ax^2 + Bx +A) = (Bx^4 + Ax^3 +...)
subst 6,
3144A -750 B = 144

[nishant.pilania]

Originally Posted by pquanta

base 6.
A =1, B = 4.
2*36 + 3*6 +1 = 91
91(Ax^2 + Bx +A) = (Bx^4 + Ax^3 +...)

how did u get to this solution?

[pquanta]

Originally Posted by nishant.pilania

whats the series
i'm not getting the relation
plz help me by explaining ur series a bit

i think the series should be (7/9)*(26/twenty eight)*(63/65)*(124/126)...
bcz avg of num and denom will give consecutive perfect cubes...
please confirm the question again...

[prabhakar85]

Originally Posted by nishant.pilania

how did u get to this solution?

The base cannot be 1,2,3,4 as there numbers are explicitly given in the question. So base has to be between 5 to 9 since it is given base less than 10.

Now 231 mulitplied by ABA gives the unit field as A which implies 231 in any base has to end in unit digit of one. From 5 to 9 one base 6 satisifies this giving 91. Hence base is 6. With Base 6 solving we can get A to be 1 and B to be 4.

[DON@IIM]

Originally Posted by prabhakar85

The base cannot be 1,2,3,4 as there numbers are explicitly given in the question. So base has to be between 5 to 9 since it is given base less than 10.

Now 231 mulitplied by ABA gives the unit field as A which implies 231 in any base has to end in unit digit of one. From 5 to 9 one base 6 satisifies this giving 91. Hence base is 6. With Base 6 solving we can get A to be 1 and B to be 4.

Hi, I agree. But to a certain extent. Tell me what's the unit digit when you multiply 136 by 424? It's 4. So it's not necessary that to get the same unit's digit in the product (in the eg. I mentioned, it's 4) one multiplicand has to end with 1 (see it has ended with 6). Isn't it?...

[prabhakar85]

Originally Posted by DON@IIM

Hi, I agree. But to a certain extent. Tell me what's the unit digit when you multiply 136 by 424? It's 4. So it's not necessary that to get the same unit's digit in the product (in the eg. I mentioned, it's 4) one multiplicand has to end with 1 (see it has ended with 6). Isn't it?...

Yeah agreed... I too thought about it... In this case 231 gives

231 in base 5 gives 70 here A cannot be zero
231 in base 6 gives 91 This satisfies.
231 in base 7 gives 126 Here A can be 4 as you say.
231 in base 8 gives 160 Here A cannot be zero
231 in base 9 gives 198 Here there is no value giving A again.

We will check for base 7 alone now. With base 7 'BA4AA' gives in base 10 as
4+28+156+2492+(7^4 * B) = will not give unit digit again to be 4. hence 7 cannot be the base.

So the above case only base 6 satisfies the given question. Please correct me if i am wrong.

[Mani_23]

solve dis question

(6666.............n units)^2 +(8888.....................n digits) is equal to??