CAT 2008: Quantitative Questions a Day 51-100 -The discussions-Brought to you by SMOT

Aarav

Quote:

Originally Posted by slam

I know its late, but it's been too long now (almost a week) and I'm rather pissed with myself for not answering the day's question on time,

p.s. its good to be back!

Feels good to see all the students who were at the start continuing to be a part of regular solvers, at least I get a lot of motivation by this

else it seems like a street, people just coming in and moving out.

vishal88arora

hey its really simple
k=a*n + 31 ............... (1)
d+k= b*n + 16 ................. (2)
d=3*k ........... (3)
using (1) and (2)
4an + 124 = bn+16
(b-4a)*n=108
this implies n is factor of 108
but n>31 acc to (1)
therefore
n is out of 36,54,108
if n=104
using (1) and (3)
d= 3*a*n + 93
then since d is largest 3 digit no.
a can be atmax 2
so, d=727
if n=54
a will be 5
so d= 903
if n=36
a=8
d=957
which is largest out of 3
so d=957
and
that 2 digit no. is 11
so answer ........... ..........(a)

implex

------------------------------------------------------
Quantitative Question # 065
------------------------------------------------------
If all palindromes (positive integers which is unchanged if you reverse the order of its digits) are written in increasing order, what is the possible number of prime values can the difference between successive palindromes take?

(1) 0 (2) 1 (3) 2 (4) 3 (5) none of these

implex

Quote:

Originally Posted by implex

------------------------------------------------------
Quantitative Question # 065
------------------------------------------------------
If all palindromes (positive integers which is unchanged if you reverse the order of its digits) are written in increasing order, what is the possible number of prime values can the difference between successive palindromes take?

(1) 0 (2) 1 (3) 2 (4) 3 (5) none of these

single digit numbers are all palindromes

for two digit numbers
10a+b=10b+a
a=b
gives us 11,22,..99
three digit numbers
100a+10b+c=100c+10b+a
=>a=c
two successive palindromes will differ by 10 or 11

for 4 digits
1000a+100b+10c+d=1000d+100c+10b+a
999(a-d)=99(c-d)
abba

so palindromes are of the form a,aa,aba,abba,abcba....
two prime numbers are 2 and 11

options 3)

implex

Quote:

Originally Posted by implex

single digit numbers are all palindromes

for two digit numbers
10a+b=10b+a
a=b
gives us 11,22,..99
three digit numbers
100a+10b+c=100c+10b+a
=>a=c
two successive palindromes will differ by 10 or 11

for 4 digits
1000a+100b+10c+d=1000d+100c+10b+a
999(a-d)=99(c-d)
abba

so palindromes are of the form a,aa,aba,abba,abcba....
two prime numbers are 2 and 11

options 3)

the fact that two successive palindromes having same no of digits where digit >2 has non prime difference
is two successive palindromes will differ only in the tens place or in the 100s place [(b+1)-b].10 so the difference will be 10 or 100 or 1000...

if the difference is other than 10,100 , it must be change in the units digit which will lead to a difference of 2

At first sight this might look vague.
But it is totally rigorous.
Check for yourself!

constant

Answer should be option 3) 2

Soln:
For even digit numbers (2n number of digits):
a1a2a3...anan...a3a2a1 would be a palindromic number.
for any successive numbers (same number of even digits) the difference would be a multiple of 10 unless the unit digit changes. The unit digit would change only in the following case:
a199...99...99a1 ----------- 1
(a1+1)00...00...00(a1+1) ---------- 2
So, 2-1 = 11
Similarly, for odd digit numbers (2n+1 number of digits)
a1a2a3...ana(n+1)an...a3a2a1 wld be palindromic number
same logic applies as for even digit numbers..
a199..999..99a1 -------- 1
(a1+1)00...000...00(a1+1) ---------- 2
so, 2-1 = 11

Now, for the even digit numbers to odd digit numbers transition and vice versa in the sequence the format would be always like
999...999...999 ---- n digits ---1
100...00...001 ----- n+1 digits ---2
2-1 = 2

So, in total 2 primes 11 and 2

Iota · 07:06 AM

QQAD : 065

Option : (3)

The difference of the successive palindromes can be formed in 2 ways:
1) keeping the unit digit same and taking the difference will always give multiple of 10.
==>p(q+1)(q+1)p- pqqp [whr pqqp represents a 4 digit number]
= 10 , simillarly can be seen for any numbers of digits.

2) increasing the unit digit and the first digit and forming the palindrome
the difference from the previous palindrome will be : 11

==> (p+1)00(p+1) - p99p = 11
same for any number of digits.

3) whr the prior palindrome consists of all 9s, then the difference with the nxt palindrome is 2.
===> 10001 - 9999 = 2
simillarly for any number of digits.

so only 2 prime numbers in the difference.

rajeev_hts

one prime number 11 comes as difference when we have 2 digit palindromes, and another diff, is 2 that is between last palindrome of n digit number and first palindrome of n+1 digit number.

like 1001 - 999 =2
101-99=2

so answer is (3)

selebratinglife

Posting just for the records..Answer is option3 that is 2(2,11)

nbangalorekar · 08:37 AM

Quote:

Originally Posted by implex

the fact that two successive palindromes having same no of digits where digit >2 has non prime difference
is two successive palindromes will differ only in the tens place or in the 100s place [(b+1)-b].10 so the difference will be 10 or 100 or 1000...

if the difference is other than 10,100 , it must be change in the units digit which will lead to a difference of 2

At first sight this might look vague.
But it is totally rigorous.
Check for yourself!

it can be proved in the foll manner:
we consider 2 cases:
case 1: when the no. of digits is odd...
take an eg of 5 digit palindrome. it will be of the form:
10001a+1010b+100c. {a,b,c can be ne of the digits with a not=0}
now successive palindrome will differ only in 'c'. so the diff will be (10^x)*c. so it will never be a prime.
case 2: when no. of digits=even
eg: 4 digits palindrome of the form
1001a+110b
so again diff among successive will be of the form (11*10^x)*b thus never a prime.
we will thus get only 2 primes: one when the difference is 11 {i.e. diff betwen 2 successive 2-digit primes}
or when diff=2 {when we move from a lower to higher digit successive prime}
cheers..