A palindrome between 1,000 and 10,000 is chosen at random. What is the

Question

A palindrome between 1,000 and 10,000 is chosen at random. What is the probability that it is divisible by 7 ?

A. 1/10
B. 1/9
C. 1/7
D. 1/6
E. 1/5

GMATinsight · Accepted Answer

Let 4 digit numbers = abba = 1000*a + 100*b+10*b + a =  1001 a + 110b

Also, 1001 = 7*11*13

i.e. 1001a will always be divisible by 7 for any non-zero value (because a can't be zero being leftmost digit) of a i.e. 9 possibilities for a

now, for 110b to be divisible by 7, b must be either 0 or 7 i.e. 2 possibilities

i.e. Total possible combinations of a and b = 9*2 = 18

Total 4 digit palindrome = 9*10*1*1 = 90 numbers

Favourable Probability = 18/90 = 1/5

Answer: Option E

Archit3110   OhsostudiousMJ

gracie · Answer

there are 90 total palindromes
9 with 00 in the middle 
and 9 with 77 in the middle
are divisible by 7
18/90=1/5
E

OhsostudiousMJ · Answer

How do you find out those numbers divisible by 7?

Archit3110 · Answer

pallindromes of 4 digits; 9*10 *1*1 ; 90 no divisible by 7 from 1,000 to 10,000 = 1428-143; 1285 P ; 90/1285 =0.07 I am going wrong somewhere

GMATinsight ; sir please advise on the solution..

OhsostudiousMJ · Answer

Lovely explanation. Thanks a lot!

gracie · Answer

hi,
double the last digit and subtract it from the remaining leading truncated number. If the result is divisible by 7, then so was the original number. Apply this rule over and over again as necessary. Example: 826. Twice 6 is 12. So take 12 from the truncated 82. Now 82-12=70. This is divisible by 7, so 826 is divisible by 7 also. 
I hope this helps,
gracie

nick1816 · Answer

We can write 4 Digit palidrome number in the form of 'abba'
a can take 9 values and b can take 10 values, hence total palidrome number possibe in that range= 9*10=90
now, if 'abba' is divisible by 7, 
bba-a=bb0 should be divisible by 7
Hence b can take only 2 value 0 or 7
Hence total 4 digit palidrome numbers which are divisible by 7= 9*2=18

OhsostudiousMJ · Answer

So, should we just try it out for all palindromes? Maybe not for all 90, but atleast for 9 palindromes, we have to try it out, right?

KarishmaB · Answer

A palindrome reads the same from front and back e.g. mom, 5665 etc. To get 4 digit palindromes, once we have the first 2 digits, the next 2 digits are already defined. Say if the first two digits are 5 and 6, next two digits are 6 ad 5. So a 4 digit palindrome is ABBA We have 9 possibilities for A (since thousands digit cannot be 0) and 10 possibilities for B. So total palindromes = 9*10*1*1 = 90 ABBA = 1000A + 100B + 10B + A = 1001A + 110B 1001 is our favourite GMAT number. Check here: https://anaprep.com/number-properties-t ... -paradigm/ It is divisible by 7, 11 and 13. So for all possible 9 values of A, 1001A is certainly divisible by 7 (since already 1001 is divisible by 7) 110 = 2*5*11 (not divisible by 7) Hence, for 110B to be divisible by 7 too, B can take only 2 values of the possible 10 i.e. 0 or 7. Hence probability of the palindrome being divisible by 7 is 2/10 = 1/5 Answer (E)

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A palindrome between 1,000 and 10,000 is chosen at random. What is the

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