For a moment. let's put '+' operator in all given boxes.
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 33.
Let 'x' be the number ahead of which '-' should be there where x can be any number from 1 to 9.
In above addition the x is already (unknown value) added so first we need to subtract it twice from LHS; first to compensate the addition operation & second for the mandatory subtraction operator that is to be placed at right place.
So for example, if it's ahead of 4 as right place then
1 + 2 + 3 + 4 - 4 - 4 + 5 + 6 + 7 + 8 + 9 = 33.
1 + 2 + 3 + (4 - 4) - 4 + 5 + 6 + 7 + 8 + 9 = 33.
For x, which is already there & can be any from 1 to 9,
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 - x - x = 33.
45 - 2x = 33
2x = 12
x = 6.
Hence, the subtraction (-) sign should be ahead of 6.
There is a ten-digit mystery number (no leading 0), represented by
ABCDEFGHIJ, where each numeral, 0 through 9, is used once.
Given the
following clues, what is the number?
1) Digit A is either a square number or a triangle number, but not both.
2) Digit B is either an even number or a cube number, but not both.
3) Digit C is either a cube number or a triangle number, but not both.
4) Digit D is either an odd number or a square number, but not both.
5) Digit E is either an odd number or a cube number, but not both.
6) Digit F is either an odd number or a triangle number, but not both.
7) Digit G is either an odd number or a prime number, but not both.
8) Digit H is either an even number or a square number, but not both.
9) Digit I is either a square number or a cube number, but not both.
10) Digit J is either a prime number or a triangle number, but not both.
11) A < B, C < D, E < F, G < H, I < J
12) A + B + C + D + E < F + G + H + I + J
Taking a look at the given clues first for identifying the mystery number ABCDEFGHIJ.
--------------------------------------------------------------------------------- 1) Digit A is either a square number or a triangle number, but not both.
2) Digit B is either an even number or a cube number, but not both.
3) Digit C is either a cube number or a triangle number, but not both.
4) Digit D is either an odd number or a square number, but not both.
5) Digit E is either an odd number or a cube number, but not both.
6) Digit F is either an odd number or a triangle number, but not both.
7) Digit G is either an odd number or a prime number, but not both.
8) Digit H is either an even number or a square number, but not both.
9) Digit I is either a square number or a cube number, but not both.
10) Digit J is either a prime number or a triangle number, but not both.
11) A < B, C < D, E < F, G < H, I < J
12) A + B + C + D + E < F + G + H + I + J
Prime numbers between 0 to 9 - 2, 3, 5, 7 Trianglenumbers between 0 to 9 - 0, 1, 3, 6 --------------------------------------------------------------------------------- STEP 1 :
A can't be 0 as no leading 0's in the mystery number ABCDEFGHIJ.
Remember both of hints can't be true at the same time. That is Digit A is either a square or a triangle number but not both. Since, 0 and 1 are square as well as triangle numbers, A can't be 0 or 1.
Similarly, 8 is even as well as cube, B can't take 8. And so on. Possible values of A - 3, 4, 6, 9. (Square/Triangle) Possible Values of B - 0, 2, 4, 6. (Even/Cube). Possible Values of C - 0, 3, 6, 8. (Cube/Triangle) Possible Values of D - 3, 4, 5, 7. (Odd/Square) Possible Values of E - 3, 5, 7, 8. (Odd/Cube) Possible Values of F - 0, 5, 6, 7, 9 (Odd/Triangle) Possible Values of G - 1, 2, 9 (Odd/Prime) Possible Values of H - 0, 2, 6, 8, 9 (Even/Square) Possible Values of I - 4, 8, 9 (Square/Cube) Possible Values ofJ- 0, 1, 2, 5, 6, 7 (Prime/Triangle)
The value ofI can't be 8 or 9 as there will be no digit left for Jwhich is greater than I.
Hence, I = 4. --------------------------------------------------------------------------------- STEP 3 : Digit A can't be 6 or 9 since no digit will be left for B>A.
As we have, I= 4, the digitA= 3 and hence B= 6. --------------------------------------------------------------------------------- STEP 4 : The only digit left for C is 0, i.e. C = 0 as C can't be 8 for C<D to be true. --------------------------------------------------------------------------------- STEP 5 : Since I = 4 itself, so digit J can't be 0, 1, 2 for I<J to be true.
Digit D and digit J take either take 5 or 7. With that the only valid digit left for E is 8 to satisfy the condition E<F. Hence E = 8.
If E = 8, then F has to be 9. So, F = 9. --------------------------------------------------------------------------------- STEP 7 : With 6, 8, 9 already taken up by other letters, the only digit left for H is 2 for G<H to be true. So, H = 3 and hence G = 1.
17 + D < 16 + J As deduced in STEP 4, D must be 5 or 7 and J must be 7 or 5. For above equation to be true, D = 5 and J = 7. --------------------------------------------------------------------------------- Conclusion :
A = 3, B = 6, C = 0, D = 5, E = 8, F = 9, G = 1, H = 2, I = 4, J = 7
Hence, the number ABCDEFGHIJ must be 3605891247.
Verifying the given hints -
1) 3 is a triangle number, but not a square number.
2) 6 is an even number, but not a cube number.
3) 0 is a cube number, but not a triangle number.
4) 5 is an odd number, but not a square number.
5) 8 is a cube number, but not an odd number.
6) 9 is an odd number, but not a triangle number.
7) 1 is an odd number, but not a prime number.
8) 2 is an even number, but not a square number.
9) 4 is a square number, but not a cube number.
10) 7 is a prime number, but not a triangle number.
11) 3 < 6, 0 < 5, 8 < 9, 1 < 2, 4 < 7
12) 3 + 6 + 0 + 5 + 8 = 22, 9 + 1 + 2 + 4 + 7 = 23, 22 < 23
There is a ten-digit mystery number (no leading 0), represented by
ABCDEFGHIJ, where each numeral, 0 through 9, is used once. Given the
following clues, what is the number?
1) Either A = B / 3 or A = G + 3.
2) Either B = I - 4 or B = E + 4.
3) Either C = J + 2 or C = F * 3.
4) Either D = G * 4 or D = E / 3.
5) Either E = J - 1 or E = D / 4.
6) Either F = B * 2 or F = A - 4.
7) Either G = F + 1 or G = I - 3.
8) Either H = A / 2 or H = C * 3.
9) Either I = H + 3 or I = D / 2.
10) Either J = H - 2 or J = C * 2.
1) Either A = B / 3 or A = G + 3. 2) Either B = I - 4 or B = E + 4. 3) Either C = J + 2 or C = F * 3. 4) Either D = G * 4 or D = E / 3. 5) Either E = J - 1 or E = D / 4. 6) Either F = B * 2 or F = A - 4. 7) Either G = F + 1 or G = I - 3. 8) Either H = A / 2 or H = C * 3. 9) Either I = H + 3 or I = D / 2. 10) Either J = H - 2 or J = C * 2.
Few digits can be eliminated for few numbers straightaway on the first go. Those digits just don't 'fit' into the both equations provided for the particular letter. 1) Either A = B / 3 or A = G + 3.
A - 0.
3) Either C = J + 2 or C = F * 3.
C - 0, 1.
4) Either D = G * 4 or D = E / 3.
D - 0, 5, 6, 7, 9
5) Either E = J - 1 or E = D / 4.
E - 9
6) Either F = B * 2 or F = A - 4.
F - 7, 9
7) Either G = F + 1 or G = I - 3.
G - 8 (since F can't be 7 & I can't be 11)
8) Either H = A / 2 or H = C * 3.
H - 0, 5, 7, 8
9) Either I = H + 3 or I = D / 2.
I - 8 (since H can't be 5 and obviously D can't be 16).
10) Either J = H - 2 or J = C * 2.
J - 3 (since H can't be 5 & C can't be 1.5), 5 (since H can't be 7 & C can't be 2.5), 9.
Now, after eliminating some digits for letters we can revise the list of digits a particular letter(s) on LHS of equation can't make because the letter(s) on RHS doesn't (don't) take some digits. For example, if J can't be 3 or 5 then C = J + 2 can't be 5 or 7. The deduction is supported by the other equation as F can't be 5/3 or 7/3. Hence, C can't be 5 or 7 for sure.
3) Either C = J + 2 or C = F * 3.
C - 0, 1,5, 7
4) Either D = G * 4 or D = E / 3.
D - 0, 5, 6, 7, 9, 3 ( since E can't be 9 & G can't be 9/4).
5) Either E = J - 1 or E = D / 4.
E - 9,4 (J can't be 5 & D can't be 16), 8 (J can't be 9 & D can't be 32).
7) Either G = F + 1 or G = I - 3.
G - 8, 0[since G can't be -1 and I can't be 3 (since if I = 3 then I = H +3 gives H = 0 but H doesn't take 0)].
------------------------------------------------------------------------------ STEP 4 : Possible values left for D are - 1, 2, 4, 8. Remember, only one of the two given hints for the particular letter has to be true. ------------------------------------------------------------------------------
STEP 4.1 : If D = 1, then D = E/3 (Hint 4) gives E = 3 & hence E = J - 1 (Hint 5) gives J = 4.
Here, D = G*4 of Hint 4 must not be true as G would be 1/4.
Similarly, E = D/4 of Hint 5 must be false and hence other hint i.e. E = J - 1 must be true.
Following the same logic as above -
If D = 1 ---> E = 3 ----> E = J - 1 ---> J = 4 ----> ---> J = H - 2 or J = C * 2.
Hence, H = 6 or C = 2.
If H = 6, then out of H = A / 2 or H = C * 3 only H = C*3 remains valid which gives C = 2.
And if C = 2, then C = J + 2 (Hint 3) gives J = 0.
So, D = 1 produces 2 different values of J as 0 and 4.
Possible values left for E are - 0, 3, 6 and 7. E = 0 ---> (Hint 5 ) ---> J = 1. But 1 is already taken by F.
E = 3 ---> (Hint 5 ) ---> J = 4 ---> (Hint 10) ---> H = 6 or invalid C = 2 H = 6 ---> (Hint 8 ) ---> invalid C = 2. The digit 2 is already taken by G, so C = 2 is invalid. Hence, E = 3 is invalid. E = 7 ---> (Hint 5 ) ---> J = 8. But 8 is already taken by G.