GMAT Questions Chat Group

How’d you get that? Please explain

I also got E, don’t know if it is correct. Total two digit and one digit numbers are 99 (1-99), that’s your sample space. For favorable outcomes, if you check for 4 - 3,4,7,8...and for 5 - 4,5,9,10....basically all multiples of 4 and 1 less than that, same for 5. Just find out how many multiples of 4 (24) and 5 (19) are there between 1 to 99. Then one less than those multiples will be also same value, 24 for 4 and 19 for 5...total adds upto 86.

So, 86/99 (E)

under exam setting, i’d solve it like this: the denominator must be a factor of 99, so eliminate B, D. Then, clearly there are more than 8 multiples of 5 between 1 and 99, so eliminate everything other than E.

that’s clever decision making.

i feel like the question is wrong, though
i just wrote a script in matlab to count such x’s manually, and the output is 69 such numbers between 1 and 99

I did the same way. so i got E as answer too.

However I was wondering as it says "Either 4 or 5", shouldn’t the unison answer be without any duplicates? In my calculation I deducted the common multiples of 4 or 5 (multiples of 20 basically" which gives 4 numbers.

So, I do get (86-4)/99= 82/99 as well. However. the closest option to that is 86/99. So I’d select that.

this double counts numbers that are divisible by both 4 and 5

it also double counts x’s such that x and x+1 both satisfy divisibility by either 4 or 5

for example, x=3, 4, 5 will be counted 4 times

as in 4 will be counted twice (srry for spammy messaging)

yeah, true. but the question is how many digits can X have. So if x=3, you’ll have 3*4 which is divisible by 4. and if x=4, you’ll have 4*5 which is divisible by 4.

the question in my mind arrives when I see the combo of " 4*5 ". Rather . cause it’s divisible by 4 and 5 both. so there could be more cases like that, sure. so 86/99 is a more simplified answer in that case.

i can see how one can arrive at that solution, but it is incorrect. Since you can arrive at a definitive answer by simply counting how many x’s satisfy the criteria given in the question (though brute force is not the best way)

also, even if we approach it this way, we would neglect to count x=99, since 99*100 is also divisible by both 4 and 5, and the less than 3 digit limitation is imposed on x, not x+1

I’m getting 79/99 - twice of (24 cases for 4, 19 cases for 5 minus 4 cases of 20) + 1 case of 99(100). Thoughts?

My approach

24 cases for x=multiple of4, 25 cases for x+1=multiple of 4, 19 cases for x=multiple of 5, 20 cases for x+1= multiple of 5.
88 cases
Minus
4 (20,40,60,80)
5 (19,39,59,79,99)
5 (4,24,44,64,84)
5 (9,29,49,69,89)

88-19= 69 cases

how-many-integers-between-1000-and-9999-have-exactly-one-pair-of-equal-348808.html


here 9*8*7*1/2 is understood and here we avoided 0 in thousands place
then we multiplied 4!/2! that is also understood

But when we multiply 4!/2! and if the number is 1104

Then the shuffle will also give me 0114

Stuck here
can someone help

Multiple of 4 : Every 2 out of 4 till 96 = 48 + 1 (from 99 * 100) = 49

Multiple of 5 : Every 2 out of 5 till 95 = 38 + 1 (from 99 * 100) = 39

Duplicate Counting : Every 4 out of 20 till 100 = 20 - 1 (Since we are not counting 100*101) = 19

Probability = (49 + 39 - 19) / (numbers from 1 to 99) = 69 / 99

when you take twice of these things to account for the x+1 factor, you double count cases where x and x+1 both satisfy the criterion. E.g. for 4, 5, you would have counted 4, 5 for 4 and 5, 6 for 5, counting 5 twice.

oops sorry, didn’t see someone already answered. I listed all the numbers out and checked. 69 is correct

Removal of duplicate cases seems quite cumbersome here. I have an alternate.

Case - 1 : Assume the first digit is equal to some other digit.

Possibilities for the first digit = 9

Choosing 1 out of the remaining 3 digits to be equal to the first = 3C1 = 3

Possibilities for the other two digits = 9 * 8 [Zero is allowed. One of the ten digits has already been taken up]

Total possibilities = 9 * 3 * 9 * 8 = 1944

Case - 2 : Assume first digit is unique.

Possibilities for the first digit = 9 [Cant be zero]

Choosing two equal digits among the remaining 3 digits = 3C2 = 3

Possibilities for assigning a number to these two digits = 9 (First digit is unique but a zero is allowed here)

Possibilities for the Fourth digit (or the second unique digit) = 8

Total = 9 * 3 * 9 * 8 = 1944

Total Cases = Case 1 + Case 2 = 1944 + 1944 = 3888

still not clear how you had 3c1 & 3c2?
i understand 9*9*8*1
How did you multiply 3

Since there can be 4 digits consider them as 3 digits ( since 2 digits are duplicate) ;

we have to fill places - - - = 9 * 9 * 8 (since 0 is not possible in first digit)