Prime Number Problem with Rot13 answer

There are 2 prime numbers between 100 and 199 such that the tens digit is a prime number, the ones digit is a prime number, and the tens and ones digit taken together are a 2 digit prime number. Find the sum of these 2 prime numbers.

Answer:

The numbers in question obviously start with "1", these are all the prime numbers between 100 and 199:

122
123*
125
127

132
133
135
137*

152
153*
155
157

172
173*
175
177.

There are more, but one or more of their digits are not prime.

Go to Rot13.com, paste the following coded message in the box, and it will instantly decipher.

Erzbir gur cevzrf jubfr graf naq barf qvtvgf ner abg cevzr:

Guvf yrnirf bar-gjragl-frira, bar-guvegl-frira, bar-svsgl-frira naq bar-friragl-guerr.

Erzbir gur cevzrf jubfr graf naq barf qvtvgf gnxra gbtrgure ner abg cevzr:

Guvf yrnirf bayl bar-guvegl-frira naq bar friragl guerr. (Gur bgure gjb-qvtvg ahzoref ner qvivfvoyr ol guerr, 57 = 3 k 19)

Nqq gurz:

Bar-guvegl-frira cyhf bar friragl guerr rdhnyf guerr uhaqerq gra.

Qb lbh frr jung “Gur Fbyire” vf qbvat? Jvgubhg hfvat n yvfg, ur dhvpxyl aneebjrq vg qbja gb 4 pubvprf, gjb bs juvpu jr pna rnfvyl ryvzvangr nf gur fhz bs gurve qvtvgf vf qvivfvoyr ol 3, urapr gubfr ahzoref ner abg cevzr.

SEND MORE MONEY PROBLEM

Solving cryptarithms[edit]

Solving a cryptarithm by hand usually involves a mix of deductions and exhaustive tests of possibilities. For instance the following sequence of deductions solves the example SEND+MORE = MONEY puzzle above (columns are numbered from right to left):

${\displaystyle {\begin{matrix}&&{\text{S}}&{\text{E}}&{\text{N}}&{\text{D}}\\+&&{\text{M}}&{\text{O}}&{\text{R}}&{\text{E}}\\\hline =&{\text{M}}&{\text{O}}&{\text{N}}&{\text{E}}&{\text{Y}}\\\end{matrix}}}$

1. From column 5, M = 1 since it is the only carry-over possible from the sum of two single digit numbers in column 4.
2. Since there is a carry in column 5, O must be less than or equal to M (from column 4). But O cannot be equal to M, so O is less than M. Therefore O = 0.
3. Since O is 1 less than M, S is either 8 or 9 depending on whether there is a carry in column 4. But if there were a carry in column 4, N would be less than or equal to O (from column 3). This is impossible since O = 0. Therefore there is no carry in column 3 and S = 9.
4. If there were no carry in column 3 then E = N, which is impossible. Therefore there is a carry and N = E + 1.
5. If there were no carry in column 2, then ( N + R ) mod 10 = E, and N = E + 1, so ( E + 1 + R ) mod 10 = E which means ( 1 + R ) mod 10 = 0, so R = 9. But S = 9, so there must be a carry in column 2 so R = 8.
6. To produce a carry in column 2, we must have D + E = 10 + Y.
7. Y is at least 2 so D + E is at least 12.
8. The only two pairs of available numbers that sum to at least 12 are (5,7) and (6,7) so either E = 7 or D = 7.
9. Since N = E + 1, E can't be 7 because then N = 8 = R so D = 7.
10. E can't be 6 because then N = 7 = D so E = 5 and N = 6.
11. D + E = 12 so Y = 2.

You can see a YouTube with a different solution method CLICK HERE or tinyurl.com/y74lvtig

ANSWERS TO "PACKET" QUESTIONS -- PAGE 1

Answers to 2017 "Algebra" Problems -- page 1

Note: this is for a group of Students in Minneapolis MN preparing for the 6th grade math challenge. It is not for the general public.

Question #1. Maria and Juan collect coins -- you have the details.

J = 25:  M = 44-25 = 19

Mgold = 19 - 8 = 11  ;  Jgold = 15 - 11 = 4 (answer)

Question #6 Julio starts a savings account -- you have the details.

let M = months.

J = 100 + 50 M   Maria = 1000 + 25 M

100 + 50 M = 1000 +25M   subtract 25 M from both ises and subtract 100 from both sides:

25 M = 900 M = 36 (answer)  Check: 100 +50 times 36 = 1900 and 1000 + 25 times 36 = 1900

Question #4: Ivy thought of a number -- you have the details.

let x be the number:

(2x - 6) divided by 2 = x - 3 : (x  - 3) times 10 + 70 =  10x + 40

10x + 40  divided by 5 = 2x + 8 = 28 solve for x, x = 10 (Answer)

Check: various stages  10 20 14 7 70 + 70 = 140 28

Question # 7  Two acute angles of a triangle are in the ratio 3:2   What are the measures of the two angles.

This is a ratio and sum problem that a student named Kai Alton had a trick for.

Guess any two angles with the correct ratio: for example, 20 and 30. Add them up and get 50.

But we are supposed to get 90. 90 divided by 50 is 1.8.

Multiply each number by 1.8 and get 36 and 54 which are in the right ratio and add up to 90.

Note: 36 is 2 times 18 and 54 is 3 times 18. 36 degrees and 54 degrees (Answer).

CRYPTARITHMETIC PROBLEM THAT IS EASY TO SOLVE

DRAFT

Crypt Arithmetic Problems That Are Easy To Solve, Revised

The assignment may seem trivial but we just want to verify that you are reading this blog. Please post your answer as a comment.

2012 6th Grade Math Challenge Problem #1
162
+XD
-----
2D7
D must equal 5 so we have:
162
+X5
-----
257
When we add X to 6, we must get 5 and a carry of 1.
So, 6 + X = 15, or X = 9
162
+95
-----
257 
Check: Do the above with a calculator and verify that it is correct. 
2012 6th Grade Math Challenge Problem #5
ON + ON + ON + ON = GO
Adding ON 4 times is the same as multiplying by 4.
ON
x4
-----
GO
Try various values for N.
N can't be zero because then, N and O would be the same.
If N is 1, O = 4 and 4 x O would be 16, generating a carry and giving a three-digit answer.
If N is 2, O = 8 and 4 x O would be 32, again generating a carry and giving a three-digit answer.
If N is 3, 4 x 3 = 12. That makes O = 2 with a carry of 1.
G = 4 x 2 + 1, or 9.
This gives:
23
x4
----
92
Check: 23 + 23 + 23 + 23 = 92
Verify this with a calculator.

About time arithmetic

TUESDAY, SEPTEMBER 24, 2013 

Repost January 19 2019

In the 5th and 6th grade math challenges, there are many problems about time.

This is was written on September 24, 2013 around the autumnal equinox.

The sun rose today at 7:02 AM and will set tonight at 7:06 PM..

It will rise tomorrow at 7:04 AM.

How long is the day September 24 and the night September 24-25?

Day and night should be approximately equal and approximately 12 hours each.

Answer below.

Yesterday, the sun rose at 7:02 AM and set at 7:06 PM. Today it rose at 7:04 AM. How long was the day on September 24 and how long was the night of September 24-25?

Note that the colon between hours and minutes is not a decimal point; time uses base 60. One each side of the colon, you can use decimal arithmetic. Working with time is a little tricky!

Day: use 24 hour clock, add 12 to PM hours. So, sun sets at 19:06. 19:06 - 7:02 = 12:04

Night: how much time between sunset and midnight? 12:00 - 7:06. You can't subtract :06 from :00, so "borrow" 60 minutes from the hour of midnight: 11:60 - 7:06 = 4:54. Now add the time until sunrise:

4:54 + 7:04 = 11:58.

The day is 12 hours and 4 minutes long and the night is 11 hours and 58 minutes long. These don't add up to 24 because we are looking at two different days. We are near the autumnal equinox so the day and night are approximately equal to 12 and approximately equal to each other.

As time goes on, days will get shorter and nights longer until the winter solstice, in December.

Today (this was September 24 2013) the sun rises at 7:04 AM and sets at 7:04 PM, so the day is exactly 12 hours.Tomorrow, the sun will rise at 7:05 AM, the night is longer than the day: prove this!

You will find the night is 11 hours and 61 minutes, so change 61 minutes to an hour and one minute