PR 4 # 29

Wacky · 2006-07-12, 01:33

Find a five digit number whose first two digits, central digit, and last two digits are perfect squares and whose square root is a prime palindrome.

Fusion_power · 2006-07-12, 03:57

I won't give the answer but will say that this is easily solved using excel.

My slightly corrupted answer is 911.

Fusion

Kees · 2006-07-12, 13:50

If the square root is a three digit prime palindrome it starts with either 1 or 3 (because otherwise its square would not be 5 digits). The highest two digitsquare is 81 so we can immediately exclude 3 which leaves 4 numbers to check: 131, 151, 181 and 191. For the last to digits of its square to be square, we need it to end on 01 or 81. Which leaves 151 and 191. We can see immediately that 151^2 would give its first two digits between 16 and 25 so our only candidate is 191. Checking gives 36481...:whistle:
Obviously, excel is faster in this kind of puzzles

Richard Cameron · 2006-07-13, 15:21

Kees

Quote:

Originally Posted by Kees

which leaves 4 numbers to check: 131, 151, 181 and 191.

I got to the answer following the same reasoning as you, but I trial multiplied the last four numbers to get the right one.

How did you get from the ten cases to four? not all are obvious non-candidates.

thanks
Richard

Fusion_power · 2006-07-13, 16:42

Here is the method I used with Excel. Its a combination of brute force and logic.

I put the following into the cells indicated:
a1 - 1
a2 - =A1+1
b1 - =A1^2
b2 - =A2^2

I pasted down the values in A2 and b2 to the next 500 rows and then copied columns a and b and pasted special values to eliminate the formulas.

The smallest 5 digit square was at 100 and the largest was at 316. I deleted all other rows keeping the remaining 217 rows.

A quick look showed that the there were several numbers that started with 2 but since all palindromes that started with 2 would end with 2 and therefore be composite, I deleted all numbers starting with 2.

I then looked at the 17 numbers that started with 3 and saw that they were all above 90,000. Since the largest 2 digit square is 81 and since 90 is larger than 81, I eliminated all numbers that started with 3 which left me with 100 numbers that started with 1.

The next step was to eliminate all non-palindrome numbers. This left me with 101, 111, 121, 131, etc.

A quick look showed that there were only 2 numbers left with 5 digit squares that started with a 2 digit square. These were 161 and 191. Since 161 squared did not end with a 2 digit square, and since 191 squared did end with a 2 digit square, I then verified the center digit in 191 was a square and had the answer.

It took less than 2 minutes to solve it this way.

Fusion

2006-07-12, 01:33	#1
Wacky Jun 2003 The Texas Hill Country 441₁₆ Posts	PR 4 # 29 Find a five digit number whose first two digits, central digit, and last two digits are perfect squares and whose square root is a prime palindrome. Last fiddled with by Wacky on 2006-07-12 at 01:34

2006-07-12, 03:57	#2
Fusion_power Aug 2003 Snicker, AL 7×137 Posts	I won't give the answer but will say that this is easily solved using excel. My slightly corrupted answer is 911. Fusion Last fiddled with by Wacky on 2006-07-12 at 04:18

2006-07-12, 13:50	#3
Kees Dec 2005 2²×7² Posts	A little dilligence If the square root is a three digit prime palindrome it starts with either 1 or 3 (because otherwise its square would not be 5 digits). The highest two digitsquare is 81 so we can immediately exclude 3 which leaves 4 numbers to check: 131, 151, 181 and 191. For the last to digits of its square to be square, we need it to end on 01 or 81. Which leaves 151 and 191. We can see immediately that 151^2 would give its first two digits between 16 and 25 so our only candidate is 191. Checking gives 36481...:whistle: Obviously, excel is faster in this kind of puzzles Last fiddled with by Kees on 2006-07-12 at 13:51

2006-07-13, 16:42	#5
Fusion_power Aug 2003 Snicker, AL 3BF₁₆ Posts	Here is the method I used with Excel. Its a combination of brute force and logic. I put the following into the cells indicated: a1 - 1 a2 - =A1+1 b1 - =A1^2 b2 - =A2^2 I pasted down the values in A2 and b2 to the next 500 rows and then copied columns a and b and pasted special values to eliminate the formulas. The smallest 5 digit square was at 100 and the largest was at 316. I deleted all other rows keeping the remaining 217 rows. A quick look showed that the there were several numbers that started with 2 but since all palindromes that started with 2 would end with 2 and therefore be composite, I deleted all numbers starting with 2. I then looked at the 17 numbers that started with 3 and saw that they were all above 90,000. Since the largest 2 digit square is 81 and since 90 is larger than 81, I eliminated all numbers that started with 3 which left me with 100 numbers that started with 1. The next step was to eliminate all non-palindrome numbers. This left me with 101, 111, 121, 131, etc. A quick look showed that there were only 2 numbers left with 5 digit squares that started with a 2 digit square. These were 161 and 191. Since 161 squared did not end with a 2 digit square, and since 191 squared did end with a 2 digit square, I then verified the center digit in 191 was a square and had the answer. It took less than 2 minutes to solve it this way. Fusion Last fiddled with by Fusion_power on 2006-07-13 at 16:43