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The 10 letters a to k leaving out stand for the digits 0 - 9 but not necessarily in that order. find which letter stands to each digit if the following sums holds:

A x B = B ----------------------F x H =CJ

B x C = AC ------------------ H x J = KJ

C x D = BC -------------------- J x K = E

D x E = CH -------------------- K x G = G

E x F = DK --------------------A x G = G


Extension

Find the missing numbers for the sum:

The 10 Letters A To K Leaving Out Stand For The Digits 0 9 But Not Necessarily In That Order Find Which Letter Stands To Each Digit If The Following Sums HoldsA class=

Sagot :

A variable holds a place in a quantity or expression and can be defined based on specific application

Part A;

A = 1, B = 3, C = 5, D = 7, E = 8, F = 9, G = 0, H = 6, J = 4, K = 2

Part B;

193, 99, 1 9 107, 99, 920, 891, 2 97, 297

The reason the above values are correct is as follows:

The given parameters are;

The letters representing the 10 numbers 0 to 9 are;

A, B, C, D, E, F, G, H, J and K

A × B = The number B...(1)

B × C = The number AC...(2)

C × D = The number BC...(3)

D × E = The number CH...(4)

E × F = The number DK...(5)

F × H = The number CJ...(6)

H × J = The number KJ...(7)

J × K = The number E...(8)

K × G = The number G...(9)

A × G = The number G...(10)

From equation (1), and (2), we have that either A = 1, or B = 1

If B = 1, then B × C = C ≠ AC

Therefore, A = 1

B × C = 1C, therefore, 1C is either 12, 14, 15, 16, or 18

From K × G = G, we have that G = 0

From the multiplication tables, we have;

4 × 6 = 24

9 × 6 = 54

Comparing with equation (6) and (7) gives;

F × H = CJ...(6)

H × J =  KJ...(7)

Let J = 4, and H = 6, therefore from J × K = E, to get a unit number, K ≤ 2

∴ K = 2, which gives;

E = 8,

From the above multiplication, C = 5, and F = 9

From which we have; B = 3

C × D = BC gives;

5 × D = 35. therefore. D = 7

Therefore, we have;

A = 1, B = 3, C = 5, D = 7, E = 8, F = 9, G = 0, H = 6,  , J = 4, K = 2

Part B

From the long division, we have that the first digit of the dividend is 1 and the first digit of the quotient is also 1, given that the 2 digit divisor number when multiplied by the first digit of quotient number gives a two digit number and that the 2 digit divisor can only divide 19X number once, we have that the possible values of 2 digit number will be larger than 95 which can be 96, 97, 98, or 99

Given that the first digit of the second to the last row is 2, we have that the last number of the quotient is 3, which gives;

9* × 1*3 = 19***

Given that the second digit of the dividend is a '9', we have that, the middle digit of the quotient is also a 9 otherwise we get a lesser value. Which gives;

9* × 193 = 19***

By checking, we also get the that the second digit of the divisor must also be a 9, to give;

99 × 193 = 19,107

We verify as follows;

193

19107 ÷ 99

[tex]{}[/tex] 99

[tex]{}[/tex]920

891

  [tex]{}[/tex]297

[tex]{}[/tex]   297

Therefore, the numbers are;

193, 99, 1 9 107, 99, 920, 891, 2 97, 297

Learn more about secret numbers here:

https://brainly.com/question/8138530

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