A palindrome is a number that reads the same forward and backward. For

Consider 7 spots.

_ _ _ _ _ _ _
0 0 0
5 5 5 5
6 6 6 6

First blank can be filled in 2 ways -5/6
2nd can be filled in 3 ways-0/5/6
3 & 4 also 3 ways.

the last 3 blanks depends on the first three so they can be filled in only 1 ways..
total - 2*3*3*3
=54

We have to form a seven-digit palindrome. Which means if we divide it in two parks, both mirror each other. In essence we only need to find the first 4 digits, the rest will be just the mirrors.
1st digit cannot be 0.
No. of ways to fill
1st slot - 2
2nd slot - 3
3rd slot 3
4th slot -4
Therefore, Total no of ways - 2*3*3*3 = 54

B

We need make palindromes of 7 digits with 3 digits that are 5, 6, 0.

Lets start filling this 7 digit palindrome from the middle.

The middle digit can be filled in 3 different ways with either of 5, 6 or 0. Now, either side of the middle digit will take same number out of 5, 6 or 0.

So, digit immediately besides can also be filled in 3 different ways, then the second and sixth place can also be filled in 3 different ways, but the extremes can be filled only in 2 ways since 0 cant go on the left end, otherwise it will be 6 digit no.

So, total possible ways: 3*3*3*2 = 54

number of options: {5,6,0} = 3
7-digit palindrome is basically: ABCDCBA
[A] cannot be 0, so 2 choices
[B] can have any 3
[C] any 3
[D] any 3
The duplicate letters need to be the same as the former, so total: 2*3*3*3=2*27=54

Answer (B).

Total digits to be used : 5,6,0
7 digit palindrome can be formed with the first digit being either 5 or 6

Identify the number of possibilities to assign a number across 7 digit places

1st digit - highest- can be filled by (5 or 6) - No of ways :2
7th digit - last digit - should be filled by the same number for palindrome, so no of ways :1
2nd digit - from highest digit- can be filled by (5 or 6 or 0) - No of ways :3
6th digit - from last digit - should be filled by the same number for palindrome, so no of ways :1
3rd digit - from highest digit- can be filled by (5 or 6 or 0) - No of ways :3
5th digit - from last digit - should be filled by the same number for palindrome, so no of ways :1
4th digit - middle digit - can be filled by (5 or 6 or 0) - No of ways :3

So total possible numbers : 2*3*3*3 = 54
IMO :B

IMO : B
This is a combination problem . As this involves one no. 0, so we can use technique, All- Minus no. with Zero,: As when zero will be first no. it cannot be 7 digit, it will become 6 digit which is the trap here.
Technique to solve palindrome questions :
Consider no. are like : abcdabc , or abababa, or bababab : You can see that we just need to see first 4 digit, rest 3 are just image of first 3 :
And 4th no. will be same as always if we come forward or backward .
So we have no. for place 1-4 : 3*3*3*3 *1*1*1 : 81 :
For image place 5,6,7th , we take it 1 only,

Now 81 is plaindrome using all no.
Considering 0 on first place we get : 1*3*3*3*1*1*1 = 27 :
81-27 = 54 :

A palindrome is a number that reads the same forward and backward. For example, 3663 and 23232 are palindromes. If 7-digit palindromes are formed using one or more of the digits 5, 6 and 0, how many such palindromes are possible?
Solution:
7 digit palindrome: so
=> 1st digit = 7th digit = 5/6 but not 0(if zero it will be 6 digit number) --> 2 choices
=> 2nd digit = 6th digit = 5/6/0 --> 3 choices
=> 3rd digit = 5th digit = 5/6/0 --> 3 choices
=> 4th digit = 5/6/0 --> 3 choices
So total number of choices = 2*3*3*3=54
A. 16
B. 54 --> correct
C. 81
D. 486
E. 729

We know that our palindrome consists of 7 digits and each digit can be 0, 5 or 6. The task is to find the number of possible palindromes.
The important condition is that palindromes read the same forward and backward i.e. the second part of the number in this case mirrors the first part and in our case. As we have 7 digits, first 3 will be mirrored by latter 3 and the 4th digit will be a unique one.
It means that latter 3 digits (5th, 6th and 7th) are dependent on the values of first 3 digits (1st, 2nd and 3rd). Taking it into account, the latter 3 digits do not play role in composing unique numbers as they in any case are the same as previous 3.
Also note that the first digit cannot be 0, otherwise there will be less than 7 digits. Thus, first digit can be only 5 or 6 i.e. has 2 values.
2nd digit can have all three values: 0, 5 and 6.
The same applies to 3rd and 4th digits.
As it was found earlier, latter three digits are dependent on the first three, and for this reason they do not add up to the total number of palindromes.
Let us now count the possible number of palindromes:
\(2*3*3*3 = 2*3^3 = 2* 27 = 54\)

Answer: B

we gave 7 digits, with 3 of them (red) are a reflection of the other 3 in the other side (green) _ _ _ _ _ _ _
so there are 4 effective slots (blue and green) that affect the possible combinations.

there is no restrictions on choosing among the three digits (0,5,6) except for the unit slot which can't be zero because this will result in 6 digit number.
so the possible outcomes = 3*3*3*2 = 54
B

We will use slot method to fill seven slots

_x _ x _x _x _ x _ x _

And means multiply

Palindrome means whatever comes in the first digit, same digit will come at the last and so on.
7 digit palindrome is to be formed with the digits 5,6 and 0.
0 cannot come at the first digit from left to right, so 2 ways to fill this position.
Second position can be filled in 3 ways; third in 3 ways; forth in 3 ways and after this fifth, sixty and seventh in 1 way each as digits will repeat itself.

2x3x3x3x3x1x1x1 = 54

B is correct.

Since this is a 7 digit number, the first digit cannot be 0 which means the last digit also cannot be 0 because this is a palindrome

With 7 digits, the number "mirrors" about the median digit which is the fourth digit. This means 3rd & 5th are the same digits, 2nd & 6th are the same digits, 1st & 7th are the same digits

So we are free to choose any of the 3 numbers for each of the second, third and fourth digits but the fifth sixth and seventh digits are restricted by our choice of the first three digits

The following shows the number of options available for each of the 7 digits

_2_ | _3_ | _3_ | _3_ | _1_ | _1_ | _1_

So the total number of such palindromes is 2*3*3*3*1*1*1 = 54

Answer is (B)

Any 7-digit palindromes can be written as 'abcdcba'.

Number of values that a can take= 2 (5 or 6)

Number of values that b, c and d can take= 3 (0, 5 or 6)

Total number of 7-digit palindromes using 0, 5 and 6 = 2*3*3*3*1*1*1= 54

IMO B

Given,

All the digits of a 7-digit palindrome are from 5,6,0 with repetitions.

To find,

Number of such palindromes possible.

Let the 7-digit number be depicted as - _ _ _ | _ | _ _ _

Approach 1: Probability approach -

Number of ways in which we can arrange the first digit = 2 (Since it can only be 5 or 6).
Number of ways in which we can arrange the second digit = 3 (Since it can be 5 or 6 or 0).
Number of ways in which we can arrange the Third digit = 3 (Since it can be 5 or 6 or 0).
Number of ways in which we can arrange the Fourth digit = 3 (Since it can be 5 or 6 or 0).
Number of ways in which we can arrange the Fifth digit = 1 (Since it has to be the same as the ten thousands digit to make it a palindrome)
Number of ways in which we can arrange the Sixth digit = 1 (Since it has to be the same as the one lakh's digit to make it a palindrome)
Number of ways in which we can arrange the Seventh digit = 1 (Since it has to be the same as the ten lakh's digit to make it a palindrome)

Now,

Total number of possible palindromes = 2 * 3 * 3 * 3 * 1 * 1 * 1 = 54

Approach 2: Combinatorics approach -

Number of ways in which we can select the first digit = 2C1 = 2 (Since it can only be from 5 or 6).
Number of ways in which we can select the second digit = 3C1 = 3 (Since it can only be from 5 or 6 or 0).
Number of ways in which we can select the Third digit = 3C1 = 3(Since it can only be from 5 or 6 or 0).
Number of ways in which we can select the Fourth digit = 3C1 = 3(Since it can only be from 5 or 6 or 0).
Number of ways in which we can select the Fifth digit = 1C1 = 1 (Since it has to be the same as the ten thousands digit to make it a palindrome)
Number of ways in which we can select the Sixth digit = 1C1 = 1 (Since it has to be the same as the one lakh's digit to make it a palindrome)
Number of ways in which we can select the Seventh digit = 1C1 = 1 (Since it has to be the same as the ten lakh's digit to make it a palindrome)

Now,

Total number of possible palindromes = 2 * 3 * 3 * 3 * 1 * 1 * 1 = 54

Answer: B

A palindrome is a number that reads the same forward and backward. For example, 3663 and 23232 are palindromes. If 7-digit palindromes are formed using one or more of the digits 5, 6 and 0, how many such palindromes are possible?


Here, the number has 7 digits and last 3 digits will depend on the first 3 digits. So effectively we have to arrange the first 4 digits only.

Things to keep in mind:
* The first digit cannot be '0', only '5' & '6' can take the first place.
* Remaining 3 places can be taken by any digit and repetition is allowed.

First place = 2 ways
Second place = 3 ways
Third place = 3 ways
Fourth place = 3 ways

Total different ways = 2*3*3*3 = 54
Total 54 different numbers are possible with given properties.

ANSWER : B

There are 7 places and 3 numbers to choose from. But, since it is a palindrome, first 3 digits will be same as last 3 digits. Further, first digit can not be zero and the fourth digit can be any of the 3 numbers.
So, effectively we have 4 places to be filled from 3 numbers in the following way,
First Digit: 2 ways (either 5 or 6)
Second Digit: 3 ways (5,6 or 0)
Third Digit: 3 ways (5,6 or 0)
Fourth Digit: 3 ways (5,6 or 0)
Total number of possible ways: 2x3x3x3= 54 ways

Ans: B

For 7 digit palindrome we need to find combinations for the first 4 digits. Last 3 digits will always need to be equal to first 3 digits, therefore cannot "control" them. Digit in the middle doesn't depend on any digits before or after. First digit has to be more than 0 therefore, 2 combinations of 5 or 6. 2nd, 3rth and 4th digits can be any of 3 digits (5,6,0):
Therefore, 2*3*3*3 = 54
Answer is B

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.