```Question 626199
problem number 1:
fatma is thinking of a 3 digit odd number. the hundreds digit is 3 times more than the units digit. the sum of the three digits is 4. what number is fatma thinking of

since the number is odd, the ones digit can only be { 1,3,5,7, or 9 }
since the hundreds digit has to be 3 times the ones digit, it can only be 3 or 9.
3 is 3 times 1
9 is 3 times 3
since the sum of the digits has to be equal to 4, we get:
hundreds digit equals 3
tens digit equals 0
ones digit equals 1
the number is 301
sum of the digits is 4
hundreds digit is 3 times units digit
number is odd

problem number 2:
who am i, i have 3 digits. i can only be divided by myself and one. the sum of my digits is 11. i am under 150.

the first digit has to be 1 since the number has to be less than 150.
the sum of all 3 digits have to be equal to 11
this means the sum of the tens digits and the ones digit has to be 10 because the hundreds digit has to be 1.
you have the following options:
119
128
137
146
that all since your number has to be below 150.
128 and 146 can't be the number because they are divisible by 2.
you are left with 119 or 137
the number has to be prime (only divisible by itself or 1 means the number is prime).
both are divisible by 1
neither is divisible by 2
neither is divisible by 3
neither by 4
neither by 5
neither by 6
neither by 7
neither by 8
neither by 9
they look to both be prime.
i checked the web for prime numbers to see if either one or both was on the list.
here's the list i got:
i know it was cheating, but i didn't feel like going through the drudgery of testing each one.
from this list, it looks like 137 is the prime number and 119 is not.
that means that 119 must be divisible by something other than 119 or 1.
since i'm basically lazy, i enlisted the help of microsoft excel to find the numbers that are divisible into 119.
the results of that analysis are shown below:
<pre>
4 (minus 2 = 2)		2 (minus 2 = 0)
number is divisible by 1 and divisible by itself so the total number of numbers
it is divisible by are the count minus 2.  the count above shows that 119 has
2 divisors other than 1 and itself, while 137 has 0 divisors other than 1 and
itself.
the actual search is below;
the first column is the divisor
the second column is the number 119 divided by the divisor in column 1.
the third column is equal to 1 if the answer is an integer and is equal to
0 if the answer is not an integer.
the fourth column is the number 137 divided by the divisor in column 1.
the fifth column is equal to 1 if the answer is an integer and is equal to
0 if the answer is not an integer.

1		119	        1		137	        1
2		59.5	        0		68.5	        0
3		39.66666667	0		45.66666667	0
4		29.75	        0		34.25	        0
5		23.8	        0		27.4	        0
6		19.83333333	0		22.83333333	0
7		17	        1		19.57142857	0
8		14.875	        0		17.125	        0
9		13.22222222	0		15.22222222	0
10		11.9	        0		13.7	        0
11		10.81818182	0		12.45454545	0
12		9.916666667	0		11.41666667	0
13		9.153846154	0		10.53846154	0
14		8.5	        0		9.785714286	0
15		7.933333333	0		9.133333333	0
16		7.4375	        0		8.5625	        0
17		7	        1		8.058823529	0
18		6.611111111	0		7.611111111	0
19		6.263157895	0		7.210526316	0
20		5.95	        0		6.85	        0
21		5.666666667	0		6.523809524	0
22		5.409090909	0		6.227272727	0
23		5.173913043	0		5.956521739	0
24		4.958333333	0		5.708333333	0
25		4.76	        0		5.48	        0
26		4.576923077	0		5.269230769	0
27		4.407407407	0		5.074074074	0
28		4.25	        0		4.892857143	0
29		4.103448276	0		4.724137931	0
30		3.966666667	0		4.566666667	0
31		3.838709677	0		4.419354839	0
32		3.71875	        0		4.28125	        0
33		3.606060606	0		4.151515152	0
34		3.5	        0		4.029411765	0
35		3.4	        0		3.914285714	0
36		3.305555556	0		3.805555556	0
37		3.216216216	0		3.702702703	0
38		3.131578947	0		3.605263158	0
39		3.051282051	0		3.512820513	0
40		2.975	        0		3.425	        0
41		2.902439024	0		3.341463415	0
42		2.833333333	0		3.261904762	0
43		2.76744186	0		3.186046512	0
44		2.704545455	0		3.113636364	0
45		2.644444444	0		3.044444444	0
46		2.586956522	0		2.97826087	0
47		2.531914894	0		2.914893617	0
48		2.479166667	0		2.854166667	0
49		2.428571429	0		2.795918367	0
50		2.38	        0		2.74	        0
51		2.333333333	0		2.68627451	0
52		2.288461538	0		2.634615385	0
53		2.245283019	0		2.58490566	0
54		2.203703704	0		2.537037037	0
55		2.163636364	0		2.490909091	0
56		2.125	        0		2.446428571	0
57		2.087719298	0		2.403508772	0
58		2.051724138	0		2.362068966	0
59		2.016949153	0		2.322033898	0
60		1.983333333	0		2.283333333	0
61		1.950819672	0		2.245901639	0
62		1.919354839	0		2.209677419	0
63		1.888888889	0		2.174603175	0
64		1.859375	0		2.140625	0
65		1.830769231	0		2.107692308	0
66		1.803030303	0		2.075757576	0
67		1.776119403	0		2.044776119	0
68		1.75	        0		2.014705882	0
69		1.724637681	0		1.985507246	0
70		1.7	        0		1.957142857	0
71		1.676056338	0		1.929577465	0
72		1.652777778	0		1.902777778	0
73		1.630136986	0		1.876712329	0
74		1.608108108	0		1.851351351	0
75		1.586666667	0		1.826666667	0
76		1.565789474	0		1.802631579	0
77		1.545454545	0		1.779220779	0
78		1.525641026	0		1.756410256	0
79		1.506329114	0		1.734177215	0
80		1.4875	        0		1.7125	        0
81		1.469135802	0		1.691358025	0
82		1.451219512	0		1.670731707	0
83		1.43373494	0		1.65060241	0
84		1.416666667	0		1.630952381	0
85		1.4	        0		1.611764706	0
86		1.38372093	0		1.593023256	0
87		1.367816092	0		1.574712644	0
88		1.352272727	0		1.556818182	0
89		1.337078652	0		1.539325843	0
90		1.322222222	0		1.522222222	0
91		1.307692308	0		1.505494505	0
92		1.293478261	0		1.489130435	0
93		1.279569892	0		1.47311828	0
94		1.265957447	0		1.457446809	0
95		1.252631579	0		1.442105263	0
96		1.239583333	0		1.427083333	0
97		1.226804124	0		1.412371134	0
98		1.214285714	0		1.397959184	0
99		1.202020202	0		1.383838384	0
100		1.19	        0		1.37	        0
101		1.178217822	0		1.356435644	0
102		1.166666667	0		1.343137255	0
103		1.155339806	0		1.330097087	0
104		1.144230769	0		1.317307692	0
105		1.133333333	0		1.304761905	0
106		1.122641509	0		1.29245283	0
107		1.112149533	0		1.280373832	0
108		1.101851852	0		1.268518519	0
109		1.091743119	0		1.256880734	0
110		1.081818182	0		1.245454545	0
111		1.072072072	0		1.234234234	0
112		1.0625 	        0		1.223214286	0
113		1.053097345	0		1.212389381	0
114		1.043859649	0		1.201754386	0
115		1.034782609	0		1.191304348	0
116		1.025862069	0		1.181034483	0
117		1.017094017	0		1.170940171	0
118		1.008474576	0		1.161016949	0
119		1	        1		1.151260504	0
120		0.991666667	0		1.141666667	0
121		0.983471074	0		1.132231405	0
122		0.975409836	0		1.12295082	0
123		0.967479675	0		1.113821138	0
124		0.959677419	0		1.10483871	0
125		0.952	        0		1.096	        0
126		0.944444444	0		1.087301587	0
127		0.937007874	0		1.078740157	0
128		0.9296875	0		1.0703125	0
129		0.92248062	0		1.062015504	0
130		0.915384615	0		1.053846154	0
131		0.908396947	0		1.045801527	0
132		0.901515152	0		1.037878788	0
133		0.894736842	0		1.030075188	0
134		0.888059701	0		1.02238806	0
135		0.881481481	0		1.014814815	0
136		0.875	        0		1.007352941	0
137		0.868613139	0		1	        1

</pre>
looks like 119 is divisible by 2 numbers.
those numbers are 7 and 17.
looks like i missed 7 in my preliminary analysis since, if i had found it then, i would have saved some additional effort down the road.
119 / 7 = 17
7 * 17 = 70 + 49 = 119.
that's it.
119 is divisible  by 7 or 17.

there was not a formula you could used other than logic.
i tried using a formula in the first problem but it didn't really help much.
it was the logic of numbers and digits and what they could be or not be that led to the solution.

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