Question 626199: my daughter is in year 6 at school and has been given maths homework which i cant help her with i have 2 questions but need to be able to show workings out and be able to understand how we got the answer.
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 off
2)who am i, i have 3 digit i can only be divived by myself and one. the sum of my digits is 11 i am under 150
Found 2 solutions by Theo, ikleyn:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! 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:
http://wiki.answers.com/Q/List_of_prime_numbers_up_to_200
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:
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
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.
your answer has to be 137.
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.
Answer by ikleyn(52921)  (Show Source):
You can put this solution on YOUR website! .
(2) who am i, i have 3 digits. i can only be divided by myself and one. the sum of my digits is 11 and i am under 150.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
This problem is nice.
As it is given, it is clear that the solution should be (a) logical and (b) not very complicated.
The logic should be accessible for a young student (a 6th grade ?)
From this point of view, it is clear that the @Theo solution does not satisfy these criterions.
So, I will present here another, more adequate solution.
So, they want you find a 3-digit prime number between 100 and 150, with the sum of digits 11.
First of all, it is clear that the first digit of this number must be 1.
Hence, the sum of two remaining digits must be 10.
Next, the last (the ones) digits can not be even number, since, otherwise, the number in this interval is not a prime.
Also, it is clear that the last digit can not be 5, since, otherwise, the number is divisible by 5.
Taking this in account, we see that the last (= the ones) digit should be 1, or 3, or 7, or 9.
If the ones digit is 1, then the tens digit is 10-1 = 9, and the number itself is 191.
This does not work, since 191 is greater than 150.
If the ones digit is 3, then the tens digit is 10-3 = 7, and the number itself is 173.
This does not work, since 173 is greater than 150.
If the ones digit is 7, then the tens digit is 10-7 = 3, and the number itself is 137.
This works, since 137 is a prime number (it is not divisible by 2, 3, 5, 7, 11,
and it is just enough to see that 137 is a prime).
If the ones digit is 9, then the tens digit is 10-9 = 1, and the number itself is 119.
This does not work, since 119 = 7*17 is not a prime number.
Thus, we analyzed all possible cases and by the method of exclusion PROVED
that there is one and only one possible answer, which is the number of  .
Solved.
--------------------------------
The solution method is a logical analysis in three steps.
First, we filter out many possible numbers based on simple divisibility properties.
Then we analyze the remaining 4 cases and exclude three of them that do not work.
Finally, we check that the remaining number satisfies all the conditions.
This way we found a UNIQUE solution to the problem.
From the point of view of complexity, this solution is ACCESSIBLE for young students of the 6th grade interested in Math.
So, it is correct, clear, accessible, adequate and instructive - all what we do expect from a true Math solution.
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