How many 3-digit numbers can be formed so that the units place of a th

Question

How many 3-digit numbers can be formed so that the units place of a three-digit number is always prime? Repetition of digits is not allowed.

Can someone explain this with the 'fill in the blanks with number of available options' method?

OA: 256

pandeyashwin · Accepted Answer

8  *  8  *  4  = 256 ways
Prime no between 0-9 : 2 , 3 , 5 , 7
Total digits between 0-9 = 10
So unit digit can be filled in 4 ways
Now 9 digits remain. First slot cannot start with 0 otherwise it won't be a 3 digit number
So first slot can be filled in 8 ways.
Middle slot can have 0 so it can be filled in 8 ways.

Princ · Answer

manavivarma

Can you update the options and OA in the question

manavivarma · Answer

Princ there were no options to this question. Have updated the OA in my post though

manavivarma · Answer

Thanks  pandeyashwin !

vishalkazone · Answer

Three Digit numbers from total 10 digit (0-10)
Restriction: Unit digit has to be prime. i.e. Either 2,3,5,or 7

Suppose three digit is A-B-C
Case 1: Unit digit 2: 
We have 1 option for C (i.e.-2)=1
We have 8 option for A ( 1,3,4,5,6,7,8,9)=8
We have another 8 option for B (except 2 and digit taken by A)=8
So total 8*8*1=64 ( for unit digit 2)

we have total 4 cases.

64*4=256

Kudos if it helps...

Princ · Answer

Thanks for updating OA

For units place, We have  4  choices i.e  2,3,5,7 
For hundreds place, We have  8  choices   (Hundreds place cannot be  0 )- 1 (Digit already used for units place)]
For Tens place, We have  8  choices   (Digit already used for Hundreds place)- 1 (Digit already used for units place)]
Total Number of 3 digits such that units place of a the number is prime number and having distinct digits: 8*8*4=256

gr8sars · Answer

But what if after filling up the units place , we fill up the tens place. And only then the hundreds place .
Total 10 digits { 0 to 9 }

In that case wouldn't it be 
4(units place = 2,3,5,7) * 9 (tens place = {0to9} minus 1 digit which has been placed at units place ) * 7(hundreds place = all digits except 0 , units place digit and tens place digit 
Therefore , 4*9*7 ?

Posted from my mobile device

vmadan10 · Answer

Hey I have used the following approach, if you have any suggestions to how I can improve my method please feel free to HMU.

I started with finding numbers which can take the units place, (2,3,5,7) 
Now we have to fill the other two places i.e. Hundreds, Tens place. We can use 9 digits to fill these places as we can't use the prime number in the units digit to fill it. 
For the hundred place we cannot have the number '0' as the resultant number will not be a 3 digit number. So we can fill the hundreds place in 8 ways.
For the tens place we have to choose from the 9 digits available to us but we have already used a non-zero number to fill the hundreds place so there are another 8 ways to fill the tens place. 
Once we tally up we get 8*8*4 (8 ways to fill the hundreds place)*(8 ways to fill the tens place)*(4 ways to fill the units place).

Regards,
Varun Madan

ashishsamant2311 · Answer

I understood your approach, mine was different and I am going wrong somewhere, Can you help me out? I am getting different solutions when I move from ten's to hundred's vs moving from hundred's to ten's
 
ten's to hundred's
----------------------------------------
starting with the constraint
for the units place - 4 ways

for ten's place = 10 ways - 1 for the Unit's place = 9 ways
for hundred's place= 10 ways - 1 way for 0 - 1 way for unit's place - 1 way for ten's place = 7

total = 7*9*4 = 252 ways.

hundred's to ten's
----------------------------------------
starting with the constraint
for the units place - 4 ways

for hundred's place= 10 ways - 1 way for 0 - 1 way for unit's = 8 ways
for ten's place = 10 ways - 1 way for hundred's - 1 way for ten's = 8 ways

total = 8*8*4 = 256 ways

I am obviously going wrong somewhere in the first method, can someone point it out?

Fdambro294 · Answer

With the "Fill-in the Slot" Type Arrangement Questions, I always try to Start from the Most Constrained Digit/"Decision" 1st

____ : _____ : _____
100s : 10s : Units

Units Digit: We have the following Options Available to Choose as Prime Digits in the Units Place ---
2 --- 3 --- 5 --- 7

(1st) Lets Start with the Scenario in which we choose 2 as the Units Digit:

___ : ____ : 2

100s Digit:

the 100s Digit can NOT have a 0 Digit (because otherwise it would NOT be a 3 Digit No.) ---- Also, because NO Repetition of Digits is allowed, we can NOT have a 2 Digit as well

Leaves ----- 8 Options

Tens Digit:

the Tens Digit CAN have the 0 Digit as an Available Option. However, the Tens Digit can NOT have 2 NOR whichever Digit was chosen for the 100s Place.

Leaves ---- 8 Options

8 * 8 * 1 = 64 Options when the Units Digit is 2

Similar Logic will apply for the other Prime Digits of: 3 --- 5 --- 7

64 + 64 + 64 + 64 = 4 * (64) = 256 Numbers

-Answer-

256

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

How many 3-digit numbers can be formed so that the units place of a th

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

How many 3-digit numbers can be formed so that the units place of a th

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards