Word problem "what number am I?"

Question

This is the word problem I cant figure out. Please help me
What number am I?
HINT #1
I am a number between 600-800
HINT #2
I am an odd number
QUESTION
I am Divisible by 3 primes < 15
Hint #1 + #2 = answer
SOURCE QUESTION
guess the number I am.
-between 600-800
-Divisible by 3 primes <15.
-Odd
-The first number and the third number add up to the middle number

Answer 1

That would be $693=3^2 \cdot 7 \cdot 11$.

Answer 2

To provide justification for @Babak's answer:

Assuming the correct wording of the problem is:

• I am a three digit number
• I am between $600$ and $800$
• I am odd
• I have exactly three prime divisors less than fifteen
• My second digit equals the sum of my first digit and third digit

Our first digit must be either $6$ or $7$. Our third digit must be small enough such that the first digit plus the third digit is less than or equal to nine. Further, the third digit must be odd. Our second digit must be the sum of the first and third digits.

This implies that the possible numbers ignoring the divisibility by primes requirement are $671, 693, 781$

Looking at the prime decomposition of each, $671 = 11\cdot 61$ only has one prime factor less than 15, $693 = 3^2\cdot 7\cdot 11$ has three prime factors less than 15, $781 = 11\cdot 71$ has only one prime factor less than 15.

Thus, the number we are searching for is indeed $693$ and there are no others satisfying the conditions.

Answer 3

The only primes less than 15 are 2,3,5,7,11 and this number is divisible by 3 of them.

====Edit. ARRGGGH!! this simply isn't true. I forgot 13!====

The number is odd so this number is not divisible by 2.

The first digit (call it $a$) plus the third digit (call it $c$) add up to the second (call it $b)

$a + c = b \le 9$ so $ c = b -a \le 9 - 6 = 3$. So last digit is not $5$. And as the number is odd the last digit is not $0$.

So the nummber is not divisible by $5$.

So the number is divisible by $3,7,11$

==== ARGH === no, it is divisible by $3,7,11$ or $3,7,13$ or $3, 11, 13$ or $7,11,13$====

So the number is divisible by $3*7*11 = 231$.

==== argh == or $3*7*13 = 273$ or $3*11*13= 429$ or $7*11*13=1001$ which is too big ====

$2*231 = 462 < 600$ and $4*231 = 924 > 800$.

So the number can only be $3*231 = 693$.

We can verify that $6+3 = 9$, that $600 < 693 < 800$, that $693$ is odd. And that $693 = 3^2*7*11$ is divisible by exactly 3 primes less than 15.

=== argh ===

$2*273 < 600$ and $4*273 > 800$ and $3*273 = 819$ doesn't work.

$429$ won't work either. Take my word for it.

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I might just delete this answer soon as it no longer is as slick and the other two answers are less calculation intensive.