How many integers k greater than 100 and less than 1000 are

When you understand it right it's simple. But you need to find an example :

Example1: 102 -> 201 and 102 + 99 = 201.
It will be okay for 102, 112, 122, 132, ..., 192 -> 10 times

So now we have that for the "100", "200", the "300", the "400", ... the "800" (8 times)

So 10 * 8 = 80

ANS : D.

Hope it's clear

u_h - h_u = 99

So reversing digits should be 1 digit different

2_1 and 1_2 (1132 and 1231)
3_2
4_3
5_4
6_5
7_6
8_7
9_8

So 10 * 8 = 80

How many integers K greater that 100 and less than 1000 are there such that if the hundreds and the units digits of K are reversed, the resulting integer is K+99?

a) 50
b) 60
c) 70
d) 80
e) 90

Ans is D.
The hundredth digit is always smaller from units digit by 1 as obtained by solving the equation 100a+10b+c+99=100c+10b+a.
So 'a' can have values from 1 to 8 and for every value of a, 'b' varies from 0 to 9. So total integers are 8×10.

Posted from my mobile device

I think the answer is D - 80 integers. My logic is as follows:

Since we are adding 99, the hundreds digit will need to be less than the ones digit if the corresponding number when the two digits flipped is to be larger. Since the change in magnitude is just shy of 100, the different between the two can only be one.

Look at an example case of 100 < X < 200

• 102 -> 201
• 112 -> 211
• ...
• 192 -> 291

What about for 800 < X < 900?

• 809 -> 908
• 819 -> 918
• ...
• 899 -> 998

For each of these samples, there are 10 numbers between the two stated bounds (the tens digit ranges from 0 to 9). Since we can't do this for 900 < X < 1,000, there are 8 sets of 10, or 80 total integers.

Hopefully that makes sense (and correct!). Let me know if I'm missing something here!

EDIT: Dang! Looks like I was beat to the punch...

GordonFreeman: What you said said is 100% correct and that was the way i solved it, but that took me lot of time, Do you feel there is any easier was like PuneetSood Said.

PuneetSood: What you said seems to be an Easy Way. Thank You!

PuneetSood's method is definitely slicker than mine. I only started my GMAT journey this week so when I approach problems like this I like to see all the moving parts. After I am more comfortable with a problem type I would opt for an approach like PuneetSood's.

Really appreciate, GordonFreeman

great approach, this is exactly the kind of help I was looking to find when I joined the gmat club community, thank you!

Let
a=Hundreds Digit
b=Tens Digit
c=Units Digit

Since we know that the hundreds digit and units digit (when reversed) is k+99,

100a+10b+c=a+10b+10c+99
99a-99c=99
99(a-c)=99
a-c=1

From here, we can assume that the hundreds digit and units digit must be 1 digit apart in order to satisfy this condition.

Only 9 options are possible:
9x8,8x7,7x6,6x5,5x4,4x3,3x2,2x1,1x0

We are also told in the question stem that we are restricted among 3 digits only which eliminates the 1x0 option. Now, there are only 8 possible options left. Since there is no restriction in the tens digits,

8 x 10 (tens digit can be from 0-9) = 80

Merging similar topics. Please refer to the discussion above.

Let the 3 digit number be xyz
If the units and hundred's digit is reversed then the number becomes zyx

Given:-
100z + 10y + x = 100x + 10y + z + 99

100z + 10y + x = 100x + 10y + z + 99

or, 99z - 99x = 99
or, 99(z-x)=99
or, z=x+1

which means z can take only 1 value i.e 1 more than x

Now in the original number xyz,
y can take 10 values 0 through 9
x can take 1-8(8 values) and for each x, z can take exactly 1 value (from 2 through 9)

so 8 (values for x and z) X 10 (values of y) = 80. Hence D

Answer is D

see the pic..

because the difference between any two reversed three-digit integers is a multiple of 99,
the least difference of 99 between integers indicates a difference of 1 between their hundreds and units digits
thus, 8 cases: 9_8, 8_7, 7_6, 6_5, 5_4, 4_3, 3_2, and 2_1, each with 10 possible tens digits,
produce 8*10=80 possible values of k
D

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