On 8th August, I saw an interesting math post on my fb timeline:

As usual, I forwarded this message to many teachers and parents groups... and finally shared this my young group of maths lovers i.e. my students, reading the same message as above, with the only difference being... I did not write the two prime numbers: 17 and 23

So naturally they started working on this problem... This is how they all worked together...

1) Since the last digit of sum is 8, the two addends can end in (1,7), (2,6), (3,5), (4,4), (9,9), (0,8) .... However square numbers cannot end in 7,2,8 and 3. Further, squares of primes cannot end in 4 as well. So the only option is (9,9)

2) The two prime numbers should be less than 30 because 30^2 = 900

3) In fact they should be less than 29, because 29^2 will be more than 818.. When I probed them for the reason, they said that the difference between 29^2 and 30^2 will be 29+30 = 57

4) Since the squares are ending in 9, so the primes will end in 7 or 3 i.e. both ending in 3, both ending in 7 or one ending in 3 and other in 7

5) So they started with the options that can be easily eliminated like --

(i) (17, 17) : Since 17^2 < 400, so their sum will be < 800

(ii) (23,23) : since 23^2 = 529, so sum will be > 1000

(iii) (13,13) and (13,17) : same reason as that of (i)

(iv) (23,3) and (23,7) : sum < 600

(v) (23,13) : Digital root of 23^2 + 13^2 = 7 + 7 = 5. But the required digital root is 8 (8+1+8)

If you see the snap, one of them did a mistake here... She multiplied the DRs of the 2 primes instead of adding the DRs of their respective squares... But this was smartly caught by other student while working on the next option i.e. (17,23)

(vi) (17,23) : They said that since this is the only option left and because my claim is correct, hence (17,23) "has to be" correct.... and we verified this using three ways

a) Estimation: 23^2 = 529 and 225 < 17^2 < 400 .. So their sum is between 850 and 930

b) Digital roots: 23^2 + 17^2 = 7 + 1 = 8 which is same as required DR

c) Actual calculation

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How would have you or your students solved this type of problem?

Regards

Rupesh Gesota

**Today's date: August 8**

**818 is the smallest palindrome that can be expressed as the sum of squares of two prime numbers. 818 = 17^2 + 23^2**As usual, I forwarded this message to many teachers and parents groups... and finally shared this my young group of maths lovers i.e. my students, reading the same message as above, with the only difference being... I did not write the two prime numbers: 17 and 23

So naturally they started working on this problem... This is how they all worked together...

1) Since the last digit of sum is 8, the two addends can end in (1,7), (2,6), (3,5), (4,4), (9,9), (0,8) .... However square numbers cannot end in 7,2,8 and 3. Further, squares of primes cannot end in 4 as well. So the only option is (9,9)

2) The two prime numbers should be less than 30 because 30^2 = 900

3) In fact they should be less than 29, because 29^2 will be more than 818.. When I probed them for the reason, they said that the difference between 29^2 and 30^2 will be 29+30 = 57

4) Since the squares are ending in 9, so the primes will end in 7 or 3 i.e. both ending in 3, both ending in 7 or one ending in 3 and other in 7

5) So they started with the options that can be easily eliminated like --

(i) (17, 17) : Since 17^2 < 400, so their sum will be < 800

(ii) (23,23) : since 23^2 = 529, so sum will be > 1000

(iii) (13,13) and (13,17) : same reason as that of (i)

(iv) (23,3) and (23,7) : sum < 600

(v) (23,13) : Digital root of 23^2 + 13^2 = 7 + 7 = 5. But the required digital root is 8 (8+1+8)

If you see the snap, one of them did a mistake here... She multiplied the DRs of the 2 primes instead of adding the DRs of their respective squares... But this was smartly caught by other student while working on the next option i.e. (17,23)

(vi) (17,23) : They said that since this is the only option left and because my claim is correct, hence (17,23) "has to be" correct.... and we verified this using three ways

a) Estimation: 23^2 = 529 and 225 < 17^2 < 400 .. So their sum is between 850 and 930

b) Digital roots: 23^2 + 17^2 = 7 + 1 = 8 which is same as required DR

c) Actual calculation

**When I asked them, as to why they did not use the estimation for (23,13) they said they did not want to add the two big square numbers....(529 + 169) :-))**----

How would have you or your students solved this type of problem?

Regards

Rupesh Gesota

**These students are from marathi-medium government school, class-7 and 8....To know more about the math enrichment program, check the website www.supportmentor.weebly.com**__PS:__
Briliance by these students is remarkable and the way they are developed by Rupesh is just superb.

ReplyDeleteLooks like a bunch of Sherlock Holmes of Mathematics are being nurtured here! While such math facts generally intimidate students/ adults, who have not had the privileged of being guided this way, these students seem to be ready to enthusiastically take on the challenge of finding the culprit numbers using their well honed deductive/detective skills... :)

ReplyDeleteThese skills will surely serve them in life too, where feeling intimidated when life throws problems at us, not thinking of all possibilities and just jumping to conclusions causes much pain...

While I could follow most of their thinking, some of their logic still eludes me...

Thankfully, I have the privilege of interacting with these children too... Next class perhaps, I'll let them teach me math... :) They do directly and indirectly teach me lots anyway... :))