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

Question

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

A. 50
B. 60
C. 70
D. 80
E. 90

Bunuel · Accepted Answer

Some wording. Anyway:

k  is 3-digit integer. Any 3 digit integer can be expressed as  100a+10b+c .

Now, we are told that  (100a+10b+c)+99=100c+10b+a  -->  99c-99a=99  -->  c-a=1  --> there are 8 pairs of  c  and  a  possible such that their difference to be 1: (9,8), (8,7), ..., and (2,1) (note that  a  can not be 0 as  k  is 3-digit integer and its hundreds digit,  a , is more than or equal to 1).

Also,  b , tens digit of  k , could tale 10 values (0, 1, ..., 9) so there are total of 8*10=80 such numbers possible.

Example:
102 - 201; 
112 - 211;
122 - 221;
...
899 - 998.

Answer: D.

pparasuraman · Answer

Not sure if this is the shortest.. But this is how I did this

There are 8 sets of integers with hundreds and units digits exchanged that satisfies k + 99.
1. 102 | 201 (satisfies k+99, where k = 102)
2. 203 | 302 (satisfies k+99, where k = 203)
3. ...
4. ...
5. ...
6. ...
7. 708 | 807
8. 809 | 908

Each set has 10 such numbers.
1. 102 | 201 (still k+99 holds good)
2. 112 | 211
3. 122 | 221
4. 132 | 231
5. ...
6. ...
7. ...
8. ...
9. 182 | 281
10. 192 | 291

Therefore, 8 sets with 10 such number in each set will give 8 x 10 = 80 integers.

alexn49 · Answer

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

saxenashobhit · Answer

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

tinku21rahu · Answer

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

PuneetSood · Answer

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

GordonFreeman · Answer

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...

tinku21rahu · Answer

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!

GordonFreeman · Answer

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.

PuneetSood · Answer

Really appreciate, GordonFreeman

Deacsonthree · Answer

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

justbequiet · Answer

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

Bunuel · Answer

Merging similar topics. Please refer to the discussion above.

romitsn · Answer

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

sahilvijay · Answer

Answer is D

see the pic..

gracie · Answer

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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