IIT JEE , AIEEE Forum - Mathematics , Algebra

doucmecomin

1)if the integers x and y are chosen at random from 1 to 100 then the probability that 7^x + 7^y is divisible by 5 is?

2)let y = 8x² -7x -9/ax² + 5x +c < 0 then a/c =?

doucmecomin

wot yaar ab tak no reply??????????????????????

sboosy

7¹ is 7 ..that is it ends with 7

7² is 49 ..ends with 9

7³ ends with 3

7⁴ ends with 1

after this it starts repeating

now comparing ending numbers

power 1 goes with 3 ..so that its divisible by 5

power 2 goes with 4 so that its divisible by 5

so if we see this way

take x to be 1

y can b 3 or 7 or 11 or ..just keep addin 4 ....until 99

so 25 numbers

take x to be 3

y can be 5 or 9 or ....97 ...so 24 numbers

in this way we get

25+24+23+ ....1

but this is for odd alone

similarly when x is 2

y can be 4 or 8 or 12 ..100 ...25 numbers

next 24 ...

so again we get

25+24+23+....1

but x and y can commute

so it is 4 (1+2+3+...25) / 100C₂

= 4 (25*26/2) / 100C₂

pls correct if i made any calculation mistake

doucmecomin

well thnkx got the process to think but ans is 1/8

LAMPARD

I dunno if this might be correct...
consider first four powers of 7...
7¹=7
7²=49
7³=343
7⁴=2401...
after this,the series of last digits goes on repeating..
Now,for 7^x+7^y,x can be chosen in 4 ways (from the 1st 4 no.s),y can be chosen in 4 ways(from 1st 4 no.s)...
so,total pairs of (x,y) is 16...but acc. to condn.,only(1,3) and (2,4) satisfy it...
so prob.=fav. cases/total cases=2/16=1/8...
I havent considered the rest of no.s coz the series keeps on repeating...so i think the prob. will still remain same...
Am i correct??

doucmecomin

ans to thik hai par fav events =(25*25)*2 n sample space is 100*100

doucmecomin

2 nd question toh koi solve karo

sboosy

ya really sorry for the mistake ...i did the sum hurriedly and thus made silly mistakes
i find that in the even case
the last favourable is 2 and not 1
so we get
hav to subtract 2*25 from the answer
that is 4*13*25 - 50 is the numerator = 1250
denominator is 100*100
so we get 1250/100*100 = 1/8

computer001

PL CHK UR SECOND QUESTION..

piit.jak

probability:

the total cases :

x and Y can take any value from the numbers 1 - 100 and any combinations.

so the total numbers possible is 100 * 100.

favourable cases:

the numbers x and Y will be of the form ,lets say

x = 4n+1 , 4n+2 ,4n+3,4n+4;

y= 4m+1 , 4m+2 ,4m+3,4m+4.

and note that m may be equak to n.

thus,

7^x+ 7^ymay take numbers in the above form.

but onl numbers in the form,

case 1 : 7⁴ⁿ⁺¹+ 7⁴ⁿ⁺³

and case 2: 7 ⁴ⁿ⁺²+ 7⁴ⁿ⁺⁴ will be divisible by 5.

now n and m can take values from 0 to 24.

a total of 25 numbers and in any combination.

thsu the total possible numbers for case 1 are 25 * 25 =625.

similarly for case 2.

thus a total favourable cases of 1250.

thus probability is

P = 1250/ 10000.

which is exactly 1/8.

konichiwa2x

Regarding your 2nd question, I am not able to find any particular value for a/c. However it comes out to be in the range:

. Is that what you were looking for?

computer001

@ konichiwa:
hey i dont think the term shud b 5x.. y will the sum consist of 2 tems 5x and -7x??? it wud have been 5/x i think..

konichiwa2x

irrespective of the values of a,c and the variable 'x'.
Substitute

to see that

Substitute

.

Substitude

and see that

From (2) and (3) we can see that,

Dividing by c (

),

Substituting

,

Quote from computer001:

@ konichiwa:
hey i dont think the term shud b 5x.. y will the sum consist of 2 tems 5x and -7x??? it wud have been 5/x i think..

why should it be 5/x? I was wondering whether it is bx. (another constant 'b')

hsbhatt

pit.jak is right for the 1st one.

7^x+7^y leaves the same remainder as 2^x+2^y.

Now a number is of the form 4k, 4k+1, 4k+2 or 4k+3.

Of this only ordered pairs of the form x=4k+2, y=4k or vice versa are favourable.

This gives 2*25*25 = 1250 fav pairs from 100*100 possibilities.

Hence 1/8

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