There are (2 + 1) ⋅ (1 + 1) ⋅ (2 + 1) = 18
factors of a.
We can make a list by running up the exponents:
{2}^{0} ⋅ {3}^{0} ⋅ {7}^{0}
{2}^{1} ⋅ {3}^{0} ⋅ {7}^{0}
{2}^{2} ⋅ {3}^{0} ⋅ {7}^{0}
{2}^{3} ⋅ {3}^{0} ⋅ {7}^{0}
{2}^{0} ⋅ {3}^{1} ⋅ {7}^{0}
{2}^{1} ⋅ {3}^{1} ⋅ {7}^{0}
{2}^{2} ⋅ {3}^{1} ⋅ {7}^{0}
{2}^{3} ⋅ {3}^{1} ⋅ {7}^{0}
{2}^{0} ⋅ {3}^{0} ⋅ {7}^{1}
{2}^{1} ⋅ {3}^{0} ⋅ {7}^{1}
{2}^{2} ⋅ {3}^{0} ⋅ {7}^{1}
{2}^{3} ⋅ {3}^{0} ⋅ {7}^{1}
{2}^{0} ⋅ {3}^{1} ⋅ {7}^{1}
{2}^{1} ⋅ {3}^{1} ⋅ {7}^{1}
{2}^{2} ⋅ {3}^{1} ⋅ {7}^{1}
{2}^{3} ⋅ {3}^{1} ⋅ {7}^{1}
{2}^{0} ⋅ {3}^{0} ⋅ {7}^{2}
{2}^{1} ⋅ {3}^{0} ⋅ {7}^{2}
{2}^{2} ⋅ {3}^{0} ⋅ {7}^{2}
{2}^{3} ⋅ {3}^{0} ⋅ {7}^{2}
{2}^{0} ⋅ {3}^{1} ⋅ {7}^{2}
{2}^{1} ⋅ {3}^{1} ⋅ {7}^{2}
{2}^{2} ⋅ {3}^{1} ⋅ {7}^{2}
{2}^{3} ⋅ {3}^{1} ⋅ {7}^{2}
Problem 13
No, it is only true that at least one prime factor cannot exceed
\sqrt{n}. Consider
2 ⋅ 47 = 94, where both 2 and 47
are prime. We have \sqrt{94} < 10
but 47 > 10 > \sqrt{94}.
But 2 < \sqrt{94}.
Problem 14
No. If n
is not prime, some of its factors may split between
b and
c. For
example, 2 ⋅ 3 = 6\mathrel{∣}18 = 2 ⋅ 9,
but 6 ∤ 2 and
6 ∤ 9. The factors of
n = 6, 2 and 3, are
split between a = 2
and b = 9.
Problem 23
Here it is true. Consider the prime factorizations of
b and
c. If
p\mathrel{∣}bc, then
p must
appear in one or both of those prime factorizations, and thus it must divide at least
one of b
and c.
Problem 24
Again, use the prime factorization of
n. Because
p and
q
are primes, they must appear in that factorization. Then
pq\mathrel{∣}n
because both appear, so you can commute products around to group
(pq)
and the rest of the factorization.
1554
is even, so divisible by 2, does not end in 5 or 0, so is not divisibleby 5, and has digits that add to 0\kern 18mu ({\rm mod}\kern 6mu 3),
so is divisible by 3.
1999
is not even, so is not divisible by 2, does not end in 5 or 0, so is notdivisible by 5, and has digits that add to 1\kern 18mu ({\rm mod}\kern 6mu 3),
so is not divisible by 3.
805is not divisible by 2, is divisible by 5, and is not divisible by3.
2450is divisible by 2, is divisible by 5, and is not divisible by 3.
Problem 2
2 and 3
2 and 5
3 and 5
2, 3, and 5
Problem 8
The missing digit must be divisible by 2. Also, the sum of the digits
must be congruent to 0 modulo 3. The non-blank digits already add to
0\kern 18mu ({\rm mod}\kern 6mu 3),
so the missing digit must be a multiple of 3. There is only one even multiple of
3 less than 10, so the missing digit must be 6.
Problem 14
Expanding the positional notation and simplifying,
abc,abc = a⋅(1{0}^{5}+1{0}^{2})+b⋅(1{0}^{4}+1{0}^{1})+c⋅(1{0}^{3}+1{0}^{0}) = (a⋅1{0}^{2}+b⋅1{0}^{1}+c)⋅(1{0}^{3}+1{0}^{0}) = abc⋅1001. Now
1001 = 7 ⋅ 11 ⋅ 13, so
abc,abc
is divisible by each of those.
Problem 15
ab - ba = a ⋅ 10 + b - (b ⋅ 10 + a) = (a - b) ⋅ 10 + (b - a).
Because 10 ≡ 1\kern 18mu ({\rm mod}\kern 6mu 9),
this becomes a - b + b - a ≡ 0\kern 18mu ({\rm mod}\kern 6mu 9).
The result always is a multiple of 9. Equivalently, we can rearrange
(a - b) ⋅ 10 + (b - a) = (a - b) ⋅ 10 + -1 ⋅ (a - b) = (a - b) ⋅ (10 - 1) = 9(a - b),
giving also which multiple of 9.
Here the difference is (a - c) ⋅ 1{0}^{2} + 0 + (c - a)
because the middle digit always cancels. Again, the result always is a
multiple of 9 and we can rearrange (a-c)⋅1{0}^{2}+0+(c-a) = (a-c)⋅1{0}^{2}+-1⋅(a-c) = (a-c)⋅(1{0}^{2}-1) = 99(a-c)
to see the result is a multiple of 99.
Take a familiar incomplete integer, -679-.
Using the expression of -679-
as N = 1{0}^{4} ⋅ {x}_{4} + {x}_{0} + 6790,
use 8\mathrel{∣}N
to find {x}_{0}.
Given that, use 9\mathrel{∣}N
to find {x}_{4}.
Now if 72 turkeys cost $-679-,
what is the total?
If 8\mathrel{∣}N then 8 divides the
last three digits, so 8\mathrel{∣}790 + {x}_{0}.
Thus 790 + {x}_{0} ≡ 0\kern 18mu ({\rm mod}\kern 6mu 8). Because
790 ≡ 6\kern 18mu ({\rm mod}\kern 6mu 8), we know that
{x}_{0} ≡ 2\kern 18mu ({\rm mod}\kern 6mu 8). The only decimal
digit satisfying {x}_{0} ≡ 2\kern 18mu ({\rm mod}\kern 6mu 8)
is \mathbf{{x}_{0} = 2}.
Now we have N = 1{0}^{4} ⋅ {x}_{4} + 6792.
For 9\mathrel{∣}N,
the sum of the digits must be zero modulo 9. Thus
{x}_{4} + 6 + 7 + 9 + 2 ≡ 0\kern 18mu ({\rm mod}\kern 6mu 9), or
{x}_{4} + 6 ≡ 0\kern 18mu ({\rm mod}\kern 6mu 9). Thus
\mathbf{{x}_{4} = 3}, and
\mathbf{N = 36792}.
(I forgot the decimal place in the problem, so these are very expensive turkeys.)
So if 72 turkeys cost $36792, each turkey costs $511. If I had remembered the decimal
place correctly, the turkeys cost $5.11 each.