Maths can be a standalone interest, but it can also be something that you can use in your everyday life and hobbies. Sometimes, it’s just part of your obsession with palindromes.
Susie Symmetry likes palindromes – strings of letters or numbers that read the same from right to left as they do from left to right. ‘Race car’ is an example of a palindrome, which may be at least part of the reason why Susie likes driving.
Susie’s new race car has a six-digit palindromic number on the licence plate. The number follows the pattern abccba, where a, b and c are digits.
Susie loves her new race car, and loves showing it off even more. When her friend Sam Skew eventually sees it, he – knowing how much she appreciates palindromes – makes it all a little bit meta.
“Your licence plate number is divisible by 11,” says Sam. He’s right. Susie’s plate number can be divided exactly by 11 with no remainder.
Susie thinks about this for a while, then rejoins, “All six-digit palindromic numbers are divisible by 11.”
Sam doesn’t like being one-upped. “Ah,” he says, “but your number is also divisible by 99.”
And then, just because they both like working with numbers, they explore all of this further.
1. Is Susie right? Are all six-digit palindromic numbers divisible by 11?
2. What do we know about the digits a, b and c if abccba is divisible by 99?
3. How many six-digit palindromic numbers are divisible by 99, if a is at least 1?
Scroll down for the solution to this week’s puzzle.
This week’s maths puzzle comes courtesy of Dr Eugene Gath, lecturer in Mathematics at the University of Limerick – who is actively involved in mathematics enrichment classes at UL and the Irish Mathematical Olympiad.
Susie Symmetry and Sam Skew sit back, both happy with their mathematical and palindromic workings. But what did they learn?
1. Yes, Susie was right
A number in the format abccba is equal to a0000a+b00b0+cc00.
This can also be written as a(100001)+b(10010)+c(1100).
1001 is equal to 1111–110, which is divisible by 11. Furthermore, 100001 is equal to 111111–11110), which is also divisible by 11.
Knowing that 100001, 10010 and 1100 are divisible by 11, Susie and Sam know that abccba must be, as well.
2. We know that a + b + c is divisible by 9
Susie and Sam know that abccba is divisible by 11. If it is divisible by 99, that means it is also divisible by 9.
A number is divisible by 9 if the sum of its digits is divisible by 9. Therefore, a+b+c+c+b+a should be divisible by 9. If this holds true, then a+b+c must be divisible by 9.
If Susie and Sam know a and b, then they can choose c, so that a+b+c is divisible by 9.
As a, b and c are digits, their sum is at most 27, so a+b+c can be either 9, 18 or 27.
Susie and Sam now have the following cases:
- If a+b is less than 9, then a+b+c has to be less than 18. Therefore, the only possible value of a+b+c from those listed above is 9, and c=9–a–b.
- If a+b is equal to 9, then c must be equal to either 0 or 9, giving us the palindromic numbers 180081, 189981, 270072, 279972, 360063, 369963, 450054, 459954, 540045, 549945, 630036, 639936, 720027, 729927, 810018, 819918.
- If a+b is greater than 9, but less than 18, then a+b+c is greater than 9 and less than 27. Therefore, the only possible value of a+b+c from the options above is 18. Hence, c=18–a–b.
- If a+b is equal to 18, then c must, once again, be equal to either 0 or 9. We therefore get the palindromic numbers 990099 and 999999.
Susie and Sam can see that, for each pair of digits ab, we can find one value of c so that abccba is palindromic and divisible by 99. For the special cases where a+b=9, or a+b=18, we get two values of c.
Thus, they get 90 six-figure palindromic numbers for the ab pairs 10, 11, 12, 13, … , 99, and some 10 extra six-figure palindromic numbers for the ab pairs 18, 27, 36, 45, 54, 63, 72, 81, 90 and 99.
All in all, Susie and Sam end up with 100 six-figure palindromic numbers.
Main image via Shutterstock
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