[implex]

1)for how many integers values of X is (2x^2-10x-4)/(x^2-4x+3) an integer?

Y=(2x^2-10x-4)/(x^2-4x+3)=(2x^2-10x-4)/(x-1)(X-3)
2 - (2x+10)/(x-1)(x-3)

clearly x!=1,3
x=2 works
x=4 works
x=-1,-5 works ( no other negative value will work )
x=7 works

after 7 we need not check as the denominator is always be larger than the numerator

hence we get 5 values

[crushkiller]

Originally Posted by implex

both are pretty easy problems and taken from testfunda I guess

anyways for the second first
cubes write them all
1,8,27,64,125,216
this shall be sufficient
125+64=216-27=189 but this pair is more than 100
but we need not try again a cute rearrangement will work for us !
hence |ABXY|=|5.4.3.6|=360
other pair is 64+27=91=216-125 if you could not think of it and |ABXY| remains same

X=(√3 + √5)^222
let Y =(√3 -√5)^222
clealry |(√3 -√5)| <1 so Y tends to 0 let us say 0.000000 ( n zeros ) 1
now X+Y is clearly an integer because the terms containing surd will get canceled

when we subtract Y from an integer it will read to a borrow and hence the digit after decimal will b 9

Y is defininitly a small number so y cant it be 0.00000 (n times ) a
where a can be 2 3 4 how can we say it will be 1 only ???

[prakharc]

Originally Posted by crushkiller

Y is defininitly a small number so y cant it be 0.00000 (n times ) a
where a can be 2 3 4 how can we say it will be 1 only ???

I believe it can be anything..Implex took 1 just for example...what we need to find is digit just after decimal......which will be 9

For eg 9-0.000a = 8.999(10-a)...so digit just after decimal is 9

[getintoiimb]

hi puys
pls help me with these qns

1)how many 2 digit positive integers are there which are one and a half times larger than the product of their digits?

ans : 1

2)for how many values of k is 12^12 the least common multiple of 6^6,8^8 and k?

ans : 25

3)what is the remaninder when 6^83+8^83 is divided by 49?

ans : 35

[prakharc]

Originally Posted by getintoiimb

hi puys
pls help me with these qns

1)how many 2 digit positive integers are there which are one and a half times larger than the product of their digits?

ans : 1

10a+b = (3/2)*ab => 20a+2b=3ab => b=20a/(3a-2)

b<=9 => 20a/(3a-2)<=9 => a>=3
It satisfies only for a=4...So 48

[prakharc]

Originally Posted by getintoiimb

hi puys
pls help me with these qns



2)for how many values of k is 12^12 the least common multiple of 6^6,8^8 and k?

ans : 25

12^12=2^24*3^12
6^6=2^6*3^6
8^8=2^24
LCM of 6^6 and 8^8= 2^24*3^6
So k will have 3^12 and any power of 2 ranging from 0-24
So 25 values possible... (3^12*2^0...........3^12*2^24)

[prakharc]

Originally Posted by getintoiimb

3)what is the remaninder when 6^83+8^83 is divided by 49?

ans : 35

(6^83 + 8^83)/49 =
14(6^82 - 6^81*8^1+6^80*8^2+....+8^82)/49 ...Now one 7 cancels out....
For odd power of 6 and 8 remainder is 6
For even power of 6 and 8 remainder is 1
So remainder when we divide by 7 = 2 *[ (1*42-6*41) mod7]=2*-1=-2
So -2 => remainder =5 ...
since we canceled one 7 earlier...Remainder when we divide by 49 = 7*5=35 Ans

[prateek7563]

y did u multiplied the LCM by 15 and subtracted by 7. please explain in detail. your answer is right!!!

[prateek7563]

Originally Posted by masoom

i Am Afraid That U Had Misses 12 In 10,11,15,22...

The Number Will Be K*lcm(10,11,12,15,22)-7...

Lcm Will Be 660...

So For Greatest Numer 660 * 15-7= 9893...

Y Did U Multiplied Lcm By 15 And Subtratced By 7

[amitkrsingh]

Originally Posted by getintoiimb

hi puys
pls help me with these qns

3)what is the remaninder when 6^83+8^83 is divided by 49?

ans : 35

EN of 49 = 42

So a^42%49=1

Let R be the remainder when 6^83/49
6R is the remainder of 6^84/49 which is 1 as per the above rule.

So 6R is a number divisible by 6 and leaves remainder 1 when divided by 49. Smallest number which satisfies this criteria is 246. Hence,

6R= 41

Similarly 8^83%49 = 43

Hence, (6^83 +8^83)%49= 84%49 = 35