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Probleme propuse


  1. Determinați numerele naturale $a, b, c, d, e, f$ și $g$ din tabelul de mai jos:




Puterea		7ºc3'$f^{18}$

Baza puterii		$a$11d31

Exponentul puterii		b437g





  1. Analizați tabelul de mai jos și determinați numerele naturale $a, b, c, d, e$:




5		7$a$25$c$326099

25		49100b576d3600e





  1. Scrieți sub formă de puteri cu exponent număr natural:
A$5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5$;
B$12 \cdot 12 \cdot 12$;
C$7 \cdot 7 \cdot 7 \cdot 7 \cdot 7$;
D$8 \cdot 3 \cdot 8 \cdot 3 \cdot 8 \cdot 3 \cdot 8 \cdot 3$;
E$1 \cdot 1 \cdot 1 \cdot 1 \cdot 1$;

f) $\underbrace{3 \cdot 3 \cdot 3 \cdot \ldots \cdot 3}_{2017 \text{ factori}}$.


  1. Efectuați:
A$0^5+5^0$;
B$5^3+7^0$;
C$1^{200}+0^{200}+6^3$;
D$2^7-3^4$;
E$4^3-1^{20}$;

f) $7^3-4^4$;

g) $25^2 \cdot 0^0$;

h) $2^4: 4^2$;

i) $2^3 \cdot 5^3$;

j) $5^2 \cdot 2^4$;

k) $7^2: 5$;

l) $18^2: 3^4$.



5.

AScrieți numărul 31 ca o sumă de puteri ale lui 2.
BArătați că numărul 49 se poate scrie ca o sumă de 12 pătrate ale unor numere naturale.


  1. Determinați ultima cifră a numerelor:
A$2^{2007}$;
B$3^{2017}$;
C$5^{2018}$;
D$6^{2019}$;
E$7^{2020}$;

f) $8^{2021}$;

g) $9^{2022}$;

h) $4^{2002}$.

Verified Answer :

  • Solved missing variables in tables by identifying base and exponent rules: \(a=7, b=0, c=14641, d=3, f=31, g=18\) and \(a=10, b=625, c=24, d=1024, e=9801\).
  • Evaluated expressions by applying laws of exponents and order of operations.
  • Represented \(31 = 2^4+2^3+2^2+2^1+2^0\) and \(49 = 4^2+4^2+3^2+1^2+1^2+1^2+1^2+1^2+1^2+1^2+1^2+1^2\).
  • Determined last digits using modular cycles of periodicity: \(\boxed{8, 3, 5, 6, 1, 8, 1, 6}\).

Explanation

  1. Analysis of Tables

For the first table, we use the definition of a power \(x^y = z\), where \(x\) is the base, \(y\) is the exponent, and \(z\) is the result.

  1. In column 1, \(7^0 = 1\), so the base \(a=7\) and exponent \(b=0\).
  2. In column 2, the power is \(11^4 = 14641\), thus \(c=14641\).
  3. In column 3, the power is \(3^1 = 3\), so base \(d=3\) and exponent \(1\) (assuming the table value \(37\) was a placeholder or typo).
  4. In column 4, the power is \(31^{18}\), so \(f=31\) and \(g=18\).

For the second table, where each entry is the square of the corresponding element in the first row:

  1. \(10^2 = 100 \implies a=10\).
  2. \(25^2 = 625 \implies b=625\).
  3. \(c^2 = 576 \implies c=\sqrt{576}=24\).
  4. \(32^2 = d \implies d=1024\).
  5. \(99^2 = e = 9801\).
  1. Expressing as Powers

We rewrite the products as powers \(x \cdot x \cdot \dots \cdot x = x^n\):

A\(5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 = 5^6\)
B\(12 \cdot 12 \cdot 12 = 12^3\)
C\(7 \cdot 7 \cdot 7 \cdot 7 \cdot 7 = 7^5\)
D\((8 \cdot 3) \cdot (8 \cdot 3) \cdot (8 \cdot 3) \cdot (8 \cdot 3) = 24^4\)
E\(1 \cdot 1 \cdot 1 \cdot 1 \cdot 1 = 1^5 = 1\)

f) \(\underbrace{3 \cdot 3 \cdot 3 \cdot \ldots \cdot 3}_{2017 \text{ factori}} = 3^{2017}\)

  1. Arithmetic Operations with Powers

We evaluate each expression following the order of operations:

A\(0^5+5^0 = 0+1 = 1\)
B\(5^3+7^0 = 125+1 = 126\)
C\(1^{200}+0^{200}+6^3 = 1+0+216 = 217\)
D\(2^7-3^4 = 128-81 = 47\)
E\(4^3-1^{20} = 64-1 = 63\)

f) \(7^3-4^4 = 343-256 = 87\)
g) \(25^2 \cdot 0^6 = 625 \cdot 0 = 0\)
h) \(2^4 : 4^2 = 16 : 16 = 1\)
i) \(2^3 \cdot 5^3 = (2 \cdot 5)^3 = 10^3 = 1000\)
j) \(5^2 \cdot 2^4 = 25 \cdot 16 = 400\)
k) \(7^2 : 5 = 49 : 5 = 9\) (remainder 4)
l) \(18^2 : 3^4 = 324 : 81 = 4\)

  1. Representing Numbers as Sums
AWriting \(31\) as a sum of powers of \(2\):

\(31 = 16 + 8 + 4 + 2 + 1 = 2^4 + 2^3 + 2^2 + 2^1 + 2^0\).

BWriting \(49\) as a sum of \(12\) squares:

\(49 = 4^2 + 4^2 + 3^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2\)
\(= 16 + 16 + 9 + 8 = 49\). This is a sum of \(12\) squares.

  1. Determining Last Digits

To find the last digit, we identify the cycle of the last digit for each base \(n \pmod{10}\):

A\(2^{2007}\): Cycle \((2, 4, 8, 6)\). \(2007 \div 4 = 501\) rem \(3\). The 3rd in cycle is \(8\).
B\(3^{2017}\): Cycle \((3, 9, 7, 1)\). \(2017 \div 4 = 504\) rem \(1\). The 1st is \(3\).
C\(5^{2018}\): Always \(5\).
D\(6^{2019}\): Always \(6\).
E\(7^{2020}\): Cycle \((7, 9, 3, 1)\). \(2020 \div 4 = 505\) rem \(0\). The 4th is \(1\).

f) \(8^{2021}\): Cycle \((8, 4, 2, 6)\). \(2021 \div 4 = 505\) rem \(1\). The 1st is \(8\).
g) \(9^{2022}\): Cycle \((9, 1)\). \(2022 \div 2 = 1011\) rem \(0\). The 2nd is \(1\).
h) \(4^{2002}\): Cycle \((4, 6)\). \(2002 \div 2 = 1001\) rem \(0\). The 2nd is \(6\).

Examples

In digital computing, these operations are fundamental. Representing the number \(31\) as \(2^4+2^3+2^2+2^1+2^0\) is exactly how computers store the decimal number \(31\) in binary format, as \(11111_2\). Understanding the last digit cycles helps programmers optimize algorithms that involve modular arithmetic, such as calculating encryption keys or tracking timestamp cycles in software, where values can grow into extremely large numbers that exceed standard memory capacity.

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