A: 333
The clocks in the first row represent 9 + 9 + 3
The calculators in the second row are each 10 (also 1234 added together is 10)
Each lightbulb in the third row is 15 (15 + 15 – 15) = 15.
Which means each ray above the lightbulb =3
So in the last row:
9 + 9 x (12 + 12 + 12)
9 + 9(36)
9 + 324
333
1, 2, 3…2020, 1^2, 2^2, 3^2…2020^2?
A: 1976.5
In order to put these into order from least to greatest, we have to figure out how many of the square(d) numbers are less than 2020:
There are 4040 numbers altogether so we need to know the 2020th and 2021st numbers and get the mean of those two numbers to find the median.
Inserting the 44 perfect squares above into the first 2020 numbers would make 2020 the 2064th number so we need to subtract 43 and 44 from 2020 to find those two numbers (the 2020th and 2021st).
By the way, the nearest perfect square under 2020 is 1936 which (luckily!) is too far away for us to think about!
2020 – 43 = 1977 and 2020 – 44 = 1976
Therefore the median of the list of numbers above is 1976.5.
Q: How many three-digit numbers are composed of three distinct digits such that one digit is the average of the other two?
A: 112 numbers
There are four numbers that end with zero that satisfy this condition: 120 240 360 and 480. Each of these numbers can only be rearranged four ways (as a leading zero is not allowed) making 16 combinations.
Each of the numbers below can be arranged into 6 distinct numbers. For example, 213 can be 123 132 213 231 312 321. Below are 16 numbers and each can be rearranged in 6 ways. Therefore there are 96 three-digit numbers that satisfy this condition that do not have a zero in it.
The total is 96 + 16 = 112.
213
315
324
417
426
435
519
528
537
546
639
648
657
759
768
897
Using perfect squares:
This is super tricky as some of the smaller perfect squares that are 3 digits repeat digits (like 121, 144, and 225). If we try 169 (whose square root is 13), then many of the two digit perfect squares cannot be used like 81, 64, 49, and 36).
For example, sqrt (169) – sqrt(25) – sqrt(4) would give 6.
If we bump up to 256 (whose square root is 16), then we can use 81 for the two digit and 9 for the single digit making it 16 – 9 – 3 = 4.
Using non-perfect squares:
Sqrt(145) – sqrt(92) – sqrt(6) = 0.00044178938368
Credit and thanks to openmiddle (if you use these in the classroom, try to modify the instructions to make them less Google-able)
Q: How many three digit perfect squares are the sum of eighteen consecutive positive integers?
A: There are 3 three digit perfect square that are the sum of eighteen consecutive positive integers: 225, 441 and 729.
If there are 18 numbers in a row, there will be 9 pairs of numbers that will add up to the same sum. Therefore we only need to check three digit perfect squares that are divisible by 9. Another condition is that the two middle numbers in the list have to add up to that perfect square divided by 9.
144 (12^2) is even and divided by 9 = 16 each pair would have to add up to 16. There are no two consecutive numbers that add to 16.
225 (15^2) divided by 9 is 25 so the two middle numbers are 12 and 13.
4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 has 12 and 13 for the middle numbers and every pair going outwards 11 and 14 10 and 15 etc adds up to 25. This list would make 9 pairs of 25 or 225.
324 (18^2) is also even like 144. 324 divided by 9 is 36 and there are no two consecutive integers that add to 36.
441 (21^2) divided by 9 is 49 so the two middle numbers are 24 and 25.
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33
576 (24^2) is also even. 576 divided by 9 is 64 and there are no two consecutive integers that add to 64.
729 (27^2) divided by 9 is 81 so the middle two numbers are 40 and 41.
32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49
900 (30^2) is also even. 900 divided by 9 is 100 and there are no two consecutive integers that add to 100.
Notice each number divisible by 9 is a multiple of 3 squared. The next multiple of 3 is 33 and 33^2 >1000. Therefore, the only three numbers are:
225, 441 and 729.
